cefer wrote: Prove that \[\sum_{cyc}(x+y)\sqrt{(z+x)(z+y)} \geq 4(xy+yz+zx),\] for all positive real numbers $x,y$ and $z$. $ LHS \ge \sum_{cyc}(x+y)(z+\sqrt{xy}) =$ $2\sum{xy}+\sum_{sym}(x+y)\sqrt{xy} \ge$ $ 2\sum{xy}+\sum_{sym}2\sqrt{xy}\sqrt{xy}=$ $4\sum{xy}=RHS $. [Moderator says: do not quote the entire post before yours]

From Cauchy-Schwartz-Bunyakovski inequality we get \[ \sqrt{\left(z+x\right)\left(z+y\right)}\geq z+\sqrt{xy}\] Then, using this one and \[\left(x+y\right)\sqrt{xy}\geq2\sqrt{xy}\sqrt{xy}=2xy\] and summing similar inequalities for other 2 summands we get desired inequality. (Seems to be very simple or I made some stupid mistake ???)

caoquyetthang our solutions are actually same. Too easy question for such competition. I forgot to add that $x=y=z$ is equality case, but it is obvious.

Just take $x+y=a^2,\ y+z=b^2,\ z+x=c^2$ (so we will have $2x=a^2+c^2-b^2$ and the analogues one) and the inequality rewrites as: $abc(a+b+c)\ge \displaystyle\sum(a^2+b^2-c^2)(a^2-b^2+c^2)\Leftrightarrow a^4+b^4+c^4+abc(a+b+c)\ge 2(a^2b^2+b^2c^2+c^2a^2)$ By Schur's Ineq for $r=2$, we'll get $a^4+b^4+c^4+abc(a+b+c)\ge \sum\limits_{cyc}a^3b+\sum\limits_{cyc}ab^3\ge 2(a^2b^2+b^2c^2+c^2a^2)$.

cefer wrote: Prove that \[\sum_{cyc}(x+y)\sqrt{(z+x)(z+y)} \geq 4(xy+yz+zx),\] for all positive real numbers $x,y$ and $z$. First note that $(x+y)(y+z)=xy+yz+zx+y^2$, so: \[3(xy+yz+zx)+(x^2+y^2+z^2)=(x+y)(y+z)+(y+z)(z+x)+(z+x)(x+y)\] Now rewrite given inequality in: \[\sum_{cyc}(x+y)\sqrt{(z+x)(z+y)}\geq\sum_{cyc}(z+x)(z+y)+\sum_{cyc}xy-\sum_{cyc}x^2\] Let $x+y=a,y+z=b,z+x=c$. Note that $a,b,c$ are sides of a triangle and note that: \[2\sum_{cyc}x^2-2\sum_{cyc}xy=\sum_{cyc}\left((x+z)-(y+z)\right)^2\] \[2\sum_{cyc}(z+x)(z+y)-2\sum_{cyc}(x+y)\sqrt{(z+x)(z+y)}=\sum_{cyc}\left(\sqrt{x+z}-\sqrt{y+z}\right)(x+y)\] Now rewrite inequality in: \[\sum_{cyc}\left(\sqrt{a}-\sqrt{b}\right)^2\left(a+b-c+2\sqrt{ab}\right)\geq 0\] Which is true since $a+b>c.$ $\square$

Rewrite the inequality as $\sum {\frac{1}{{\sqrt {\left( {x + y} \right)\left( {x + z} \right)} }}} \ge \frac{{4\left( {xy + yz + zx} \right)}}{{\left( {x + y} \right)\left( {y + z} \right)\left( {z + x} \right)}}$ Using the familiar inequality $\left( {x + y} \right)\left( {y + z} \right)\left( {z + x} \right) \ge \frac{8}{9}\left( {xy + yz + zx} \right)\left( {x + y + z} \right)$, it reduces to prove $\sum {\frac{1}{{\sqrt {\left( {x + y} \right)\left( {x + z} \right)} }}} \ge \frac{9}{{2\left( {x + y + z} \right)}}$ Applying Cauchy and Cauchy Schwarz inequality we get $\sum {\frac{1}{{\sqrt {\left( {x + y} \right)\left( {x + z} \right)} }}} \ge \sum {\frac{2}{{x + y + 2z}}} \ge \frac{9}{{2\left( {x + y + z} \right)}}$ Complete proof. Equality holds when $x=y=z$.

Also $A\ge G$ on the whole expression + $9(x+y)(y+z)(x+z)\ge 8(x+y+z)(xy+yz+xz)$ does the trick.

Method 1. Use Cauchy Schwarz to get $\sqrt{(z+x)(z+y)}\geqslant z+\sqrt{xy}$. Hence $\sum_{cyc}(x+y)\sqrt{(z+x)(z+y)} \geqslant \sum_{cyc}(x+y)(z+\sqrt{xy}) =\sum_{cyc}z(x+y) +\sum_{cyc}(x+y)\sqrt{xy}$. Notice that by AM-GM $x+y \geqslant 2\sqrt{xy}$, which gives $\sum_{cyc}(x+y)\sqrt{xy} \geqslant \sum_{cyc}2xy$. Therefore \[ LHS \geqslant \sum_{cyc}z(x+y) +\sum_{cyc}(x+y)\sqrt{xy} \geqslant 2(xy+yz+zx) +2(xy+yz+zx)=4(xy+yz+zx) \] Method 2. Without loss of generality, let $xy+yz+zx=1$. Then we have to prove that \[ \sum_{cyc}(x+y)\sqrt{z^2 +1} \geqslant 4 \] Observe that $f(x)=\sqrt{x^2 +1}$ is a convex function of $x$. We proceed by Jensen's inequality: \[ 2(x+y+z)\sum_{cyc}\frac{x+y}{2(x+y+z)} \sqrt{z^2 +1} \geqslant 2(x+y+z) \sqrt{\frac{1}{(x+y+z)^2}+1} \] Set $p=x+y+z$. We want to prove that $2p\sqrt{\frac{1+p^2}{p^2}}= 2\sqrt{1+p^2} \geqslant 4$. By AM-GM $p^2 =(x+y+z)^2 \geqslant 3(xy+yz+zx)=3$, hence $2\sqrt{1+p^2} \geqslant 2\cdot 2 =4$. Done.

\[ \begin{gathered} (x - y)^2 \geqslant 0 \hfill \\ \hfill \\ \Leftrightarrow (xy + yz + zx)(x^2 + y^2 ) \geqslant (xy + yz + zx)2xy \hfill \\ \hfill \\ \Leftrightarrow (xy + yz + zx)(x^2 + y^2 ) + (xy + yz + zx)^2 + x^2 y^2 \geqslant \hfill \\ (xy + yz + zx)2xy + (xy + yz + zx)^2 + x^2 y^2 \hfill \\ \hfill \\ \Leftrightarrow (xy + yz + zx)(x^2 + y^2 ) + (xy + yz + zx)^2 + x^2 y^2 \geqslant \hfill \\ (xy + (xy + yz + zx))^2 \hfill \\ \hfill \\ \Leftrightarrow (xy + yz + zx)(x^2 + xy + yz + zx) + y^2 (x^2 + xy + yz + zx) \geqslant \hfill \\ (2xy + yz + zx)^2 \hfill \\ \hfill \\ \Leftrightarrow (x^2 + xy + yz + zx)(y^2 + xy + yz + zx) \geqslant \hfill \\ (2xy + yz + zx)^2 \hfill \\ \hfill \\ \Leftrightarrow (x + y)(x + z)(x + y)(y + z) \geqslant (2xy + yz + zx)^2 \hfill \\ \hfill \\ \Leftrightarrow (x + y)\sqrt {(x + z)(y + z)} \geqslant (2xy + yz + zx) \hfill \\ \hfill \\ \Rightarrow \sum\limits_{cyc} {(x + y)\sqrt {(x + z)(y + z)} } \geqslant 4(xy + yz + zx) \hfill \\ \end{gathered} \]

Let $a,b,c$ be the sides of a triangle such that $x=b+c-a, y=a+c-b, z=a+b-c$ then we will have : $\sqrt{abc}(\sqrt{a}+\sqrt{b}+\sqrt{c})+a^{2}+b^{2}+c^{2}\ge 2(ab+bc+ac)$ which is equivalent with : $(a-b)^{2}+(b-c)^{2}+(a-c)^{2}-a(\sqrt{c}-\sqrt{b})^{2}+b(\sqrt{a}-\sqrt{c})^{2})+c(\sqrt{a}-\sqrt{b})^{2}\ge 0$ factorizing we get the following which is true by triangle inequality: $\sum{(\sqrt{a}-\sqrt{b})^{2}(\sqrt{a}+\sqrt{b}+\sqrt{c})(\sqrt{b}+\sqrt{c}-\sqrt{a})}\ge0$ If $a,b,c$ can be the sides of a triangle then $\sqrt{a},\sqrt{b},\sqrt{c}$ can also be triangle sides

The problem is equivalent to a problem of the Romanian TST 2001: http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2148433 The very form of the problem statement is actually given in Akiyama's post..

This suddenly came to me at midnight on the competition day (I got 0 in the exam): Do normal substitutions of a = root(y+z) and so on and prove that a,b,c satisfy triangle inequality, and then \[LHS=abc(a+b+c)=2abc\times s\] \[RHS=16\times Area^2\] where s is the semi-perimeter. But this easily cancels using well known formulae for the Area of triangle made by sides a,b,c. \[RHS=16\times \frac{abc}{4R} \times rs\] This cancels down nicely to Euler's inequality: \[R\geq 2r\]

\[ \sum_{cyc}(x+y)\sqrt{(z+x)(z+y)}\geq 4(xy+yz+zx), \] And sub in \[a=\sqrt{x+y}\ , b=\sqrt{x+z}\ , c=\sqrt{y+z}\] Then you get \[abc(a+b+c)\geq(a^{2}+b^{2}-c^{2})(a^{2}-b^{2}+c^{2})+(a^{2}-b^{2}+c^{2})(b^{2}+c^{2}-a^{2})+(b^{2}+c^{2}-a^{2})(a^{2}+b^{2}-c^{2})\] Note that the triangle inequality holds for a,b and c as a, b, c, x, y, z are all positive so \[a^{2}+b^{2}>c^{2}\Rightarrow\ a+b>c\] and similarly for others. Thus dividing by abc we get: \[a+b+c\geq\ a\cos{B}\cos{C}+b\cos{C}\cos{A}+c\cos{A}\cos{B}\] which is trivial as cos(x) is at most 1 for all real x

Frill, when does equality hold at your inequality? Also i think you need to prove that if with $a,b,c$ you can form a triangle, then you can also form a triangle with side lengths $\sqrt{a},\sqrt{b},\sqrt{c}$

Sorry about my last post, I was writing utter rubbish it is rather more complicated than I originally thought!: \[ \sum_{cyc}(x+y)\sqrt{(z+x)(z+y)}\geq 4(xy+yz+zx), \] And sub in \[a=\sqrt{x+y}\ , b=\sqrt{x+z}\ , c=\sqrt{y+z}\] Then you get \[abc(a+b+c)\geq(a^{2}+b^{2}-c^{2})(a^{2}-b^{2}+c^{2})+(a^{2}-b^{2}+c^{2})(b^{2}+c^{2}-a^{2})+(b^{2}+c^{2}-a^{2})(a^{2}+b^{2}-c^{2})\] Note that the triangle inequality holds for a,b and c as a, b, c, x, y, z are all positive so \[a^{2}+b^{2}>c^{2}\Rightarrow\ a+b>c\] and similarly for others. Furthermore as $a^{2}+b^{2}>c^{2}$ it is an acute-angle triangle. Thus dividing by $2abc$ we get: \[a+b+c\geq\sum_{cyc}4a\cos{B}\cos{C} = 4R\sum_{cyc}2\cos{A}\cos{B}\sin{C} = 4R\sum_{cyc}\cos{A}(\sin{B}\cos{C}+\sin{C}\cos{B}) = 4R\sum_{cyc}\cos{A}\sin{(B+C)} = 4R\sum_{cyc}\sin{A}\cos{A} = 2R\sum_{cyc}\sin{2A}\] Now the RHS is twice the perimeter of the orthic $\triangle DEF$ of $\triangle ABC$ ($EF = R\sin{2A}$ etc.). Thus we are left to prove that the perimeter of $\triangle ABC$ is greater than twice the perimeter of $\triangle DEF$. To do this you prove that $BC\geq DE+DF$ and cyclically sum it (this was British Mathematical Olympiad round 1 problem 6 this year). This is the same as proving $BD+DC\geq DE+DF$ i.e. $\sin{B}\cos{C}+\sin{C}\cos{B}\geq \sin{B}\cos{B}+\sin{C}\cos{C}$ which falls out by using the rearrangement inequality and assuming WLOG $B\geq C$ with equality iff $A=B=C$ i.e. $x=y=z$

Actually, $a^2, b^2,c^2$ are your triangle sides(as you wrote, $a=\sqrt{x+y}$ which is equivalent to $a^2=x+y$) so this does not mean that the triangle is acute, unless you prove that you can form a triangle with sidelengths $(\sqrt{x+y},\sqrt{x+z},\sqrt{z+y})=(a,b,c)$ Also i was asking when does equality hold at: $ a+b+c\geq\ a\cos{B}\cos{C}+b\cos{C}\cos{A}+c\cos{A}\cos{B} $

No my triangle sides are $a,b,c$ as that is the triangle I use later on in the proof I was just using the fact that $a^{2}+b^{2}>c^{2}$ to prove that $a^2+b^2 +2ab>c^2$ so $a+b>c$. The other inequality I had was wrong and it would have equality iff $\cos{A}=\cos{B}=\cos{C}=1$ which of course can't be valid else the triangle doesn't exist.

$x+y=c^2$ $y+z=a^2$ $x+z=b^2$ Then we have to prove that: $a^4+b^4+c^4+abc(a+b+c)\geq2(a^2b^2+b^2c^2+c^2a^2)$ From Shur inequality $a^2(a-b)(a-c)+b^2(b-a)(b-c)+c^2(c-a)(c-a)\geq0$ $a^4+b^4+c^4+abc(a+b+c)\geq(a^3b+a^3c+b^3a+b^3c+c^3a+c^3b)\geq2(a^2b^2+b^2c^2+c^2a^2)$

We do a trivial use of holder inequality . Let us call this expression in LHS as $S$ . Now , by holder inequality \[ (S)^2((x+z)^2+(y+x)^2+(z+x)^2) \ge (x^2+y^2+z^2+ 3(xy+yz+zx))^3 \] \[ 2S^2(x^2+y^2+z^2+xy+yz+xz) \ge (x^2+y^2+z^2+3(xy+yz+zx))^3 \] Thus it suffice to show that , \[ S^2 \ge \frac{(x^2+y^2+z^2+3(xy+yz+zx))^3}{2(x^2+y^2+z^2+xy+yz+zx)} \ge 16(xy+yz+zx)^2 \] Now , lets take $xy+yz+zx=a$ and let $x^2+y^2+z^2=at$ . Clearly $t \ge 1 $ Thus the inequality boils down to , $ (t+3)^3 \ge 32(t+1) $ which is equivalent to $t^3+27+9t^2+27t -32t-32 \ge 0$ which is true since $(t-1)(t^2+10t+5) \ge 0$ (since $t \ge 1$ ) . Equality holds when $t=1$ or $x=y=z$ $\Box$

The following inequality is also true

luofangxiang wrote: The following inequality is also true Let $a,b,c\geq0$, $\sum_{cyc}ab=3$ ; Prove: $\sum_{cyc}\sqrt{(a^2+3)(b^2+3)}\geq3\sum_{cyc}a+3$ We need to prove that $\sum_{cyc}(a+b)\sqrt{(a+c)(b+c)}\geq(a+b+c)\sqrt{3(ab+ac+bc)}+ab+ac+bc$ or $\sum_{cyc}(2a^2+4ab)-2(a+b+c)\sqrt{3(ab+ac+bc)}\geq\sum_{cyc}(a+b)\left(a+b+2c-2\sqrt{(a+c)(b+c)}\right)$ or $2(a+b+c)\left(a+b+c-\sqrt{3(ab+ac+bc)}\right)\geq\sum_{cyc}(a+b)\left(\sqrt{a+c}-\sqrt{b+c}\right)^2$ or $\sum_{cyc}(a-b)^2\left(\frac{a+b+c}{a+b+c+\sqrt{3(ab+ac+bc)}}-\frac{a+b}{\left(\sqrt{a+c}+\sqrt{b+c}\right)^2}\right)\geq0$, which is true because $\frac{a+b+c}{a+b+c+\sqrt{3(ab+ac+bc)}}-\frac{a+b}{\left(\sqrt{a+c}+\sqrt{b+c}\right)^2}=$ $=\frac{a+b+c}{a+b+c+\sqrt{3(ab+ac+bc)}}-\frac{a+b}{a+b+2c+2\sqrt{(a+c)(b+c)}}\geq$ $\geq\frac{a+b+c}{a+b+c+\sqrt{3(ab+ac+bc)}}-\frac{a+b+c}{a+b+c+2\sqrt{ab+ac+bc}}\geq0$. Done!

Nice，arqady

\[ \left( y+z \right) \sqrt { \left( z+x \right) \left( x+y \right) }+ \left( z+x \right) \sqrt { \left( x+y \right) \left( y+z \right) }+ \left( x+y \right) \sqrt { \left( y+z \right) \left( z+x \right) }\]\[\geq { \frac {288}{125}}\,{\frac { \left( yz+ \left( z+x \right) \left( x+y \right) \right) \left( zx+ \left( x+y \right) \left( y+z \right) \right) \left( xy+ \left( y+z \right) \left( z+x \right) \right) } { \left( y+z \right) \left( z+x \right) \left( x+y \right) \left( x +y+z \right) }}\geq 4(xy+yz+zx)\] \[{\frac {9}{16}}\,{\frac { \left( 2\,y+z+x \right) \left( 2\,z+x+y \right) \left( z+2\,x+y \right) }{x+y+z}}\geq \left( y+z \right) \sqrt { \left( z+x \right) \left( x+y \right) }+ \left( z+x \right) \sqrt { \left( x+y \right) \left( y+z \right) }+ \left( x+y \right) \sqrt { \left( y+z \right) \left( z+x \right) }\]

$ \left( y+z \right) \sqrt { \left( z+x \right) \left( x+y \right) }+ \left( z+x \right) \sqrt { \left( x+y \right) \left( y+z \right) }+ \left( x+y \right) \sqrt { \left( y+z \right) \left( z+x \right) }-{ \frac {288}{125}}\, \left( {\frac {yz}{y+z}}+{\frac { \left( z+x \right) \left( x+y \right) }{y+z}} \right) \left( {\frac {zx}{z+x}} +{\frac { \left( x+y \right) \left( y+z \right) }{z+x}} \right) \left( {\frac {xy}{x+y}}+{\frac { \left( y+z \right) \left( z+x \right) }{x+y}} \right) \left( x+y+z \right) ^{-1}\geq 0. $ $ \left( y+z \right) \sqrt { \left( z+x \right) \left( x+y \right) }+ \left( z+x \right) \sqrt { \left( x+y \right) \left( y+z \right) }+ \left( x+y \right) \sqrt { \left( y+z \right) \left( z+x \right) }-4 \,{\frac { \left( x+y+z \right) \left( x{y}^{2}+y{z}^{2}+{x}^{2}z \right) }{{x}^{2}+{y}^{2}+{z}^{2}}}\geq 0. $ $\left( y+z \right) \sqrt { \left( z+x \right) \left( x+y \right) }+ \left( z+x \right) \sqrt { \left( x+y \right) \left( y+z \right) }+ \left( x+y \right) \sqrt { \left( y+z \right) \left( z+x \right) }-6 \, \left( {\frac { \left( y+z \right) ^{2}}{2\,{x}^{2}+yz}}+{\frac { \left( z+x \right) ^{2}}{2\,{y}^{2}+zx}}+{\frac { \left( x+y \right) ^{2}}{2\,{z}^{2}+xy}}+2\,{\frac {xy+zx+yz}{{x}^{2}+{y}^{2}+{z}^{2}}} \right) xyz \left( x+y+z \right) ^{-1}\geq 0 $ $- \left( y+z \right) \sqrt { \left( z+x \right) \left( x+y \right) }- \left( z+x \right) \sqrt { \left( x+y \right) \left( y+z \right) }- \left( x+y \right) \sqrt { \left( y+z \right) \left( z+x \right) }+{ \frac {9}{16}}\,{\frac { \left( 2\,y+z+x \right) \left( 2\,z+x+y \right) \left( z+2\,x+y \right) }{x+y+z}}\geq 0 $ $-12\,{\frac {\sqrt {xy}{z}^{2}}{x+y+z}}-12\,{\frac {\sqrt {yz}{x}^{2}} {x+y+z}}-12\,{\frac {\sqrt {zx}{y}^{2}}{x+y+z}}+ \left( y+z \right) \sqrt { \left( z+x \right) \left( x+y \right) }+ \left( z+x \right) \sqrt { \left( x+y \right) \left( y+z \right) }+ \left( x+y \right) \sqrt { \left( y+z \right) \left( z+x \right) }\geq 0 $ $\left( y+z \right) \sqrt { \left( z+x \right) \left( x+y \right) }+ \left( z+x \right) \sqrt { \left( x+y \right) \left( y+z \right) }+ \left( x+y \right) \sqrt { \left( y+z \right) \left( z+x \right) }-3 \,{\frac {{x}^{2} \left( y+z \right) ^{3}}{ \left( x+y+z \right) \left( {y}^{2}+{z}^{2} \right) }}-3\,{\frac {{y}^{2} \left( z+x \right) ^{3}}{ \left( x+y+z \right) \left( {z}^{2}+{x}^{2} \right) } }-3\,{\frac {{z}^{2} \left( x+y \right) ^{3}}{ \left( x+y+z \right) \left( {x}^{2}+{y}^{2} \right) }}\geq 0 $ $ \left( y+z \right) \sqrt { \left( z+x \right) \left( x+y \right) }+ \left( z+x \right) \sqrt { \left( x+y \right) \left( y+z \right) }+ \left( x+y \right) \sqrt { \left( y+z \right) \left( z+x \right) }-6 \,{\frac {{x}^{2} \left( y+z \right) ^{3}}{ \left( z+x \right) \left( x+y+z \right) \left( x+y \right) }}-6\,{\frac {{y}^{2} \left( z+x \right) ^{3}}{ \left( x+y \right) \left( x+y+z \right) \left( y+z \right) }}-6\,{\frac {{z}^{2} \left( x+y \right) ^{3}}{ \left( y+z \right) \left( x+y+z \right) \left( z+x \right) }}\geq 0 $ $ \left( y+z \right) \sqrt { \left( z+x \right) \left( x+y \right) }+ \left( z+x \right) \sqrt { \left( x+y \right) \left( y+z \right) }+ \left( x+y \right) \sqrt { \left( y+z \right) \left( z+x \right) }-6 \,{\frac { \left( x+y \right) ^{2}xyz}{ \left( x+y+z \right) \left( { z}^{2}+xy \right) }}-6\,{\frac { \left( y+z \right) ^{2}xyz}{ \left( x +y+z \right) \left( {x}^{2}+yz \right) }}-6\,{\frac { \left( z+x \right) ^{2}xyz}{ \left( x+y+z \right) \left( {y}^{2}+zx \right) }}\geq 0 $ $ \left( y+z \right) \sqrt { \left( z+x \right) \left( x+y \right) }+ \left( z+x \right) \sqrt { \left( x+y \right) \left( y+z \right) }+ \left( x+y \right) \sqrt { \left( y+z \right) \left( z+x \right) }- 12\,{\frac { \left( x+y \right) ^{2}xyz}{ \left( y+z \right) \left( x +y+z \right) \left( z+x \right) }}-12\,{\frac { \left( y+z \right) ^{ 2}xyz}{ \left( z+x \right) \left( x+y+z \right) \left( x+y \right) } }-12\,{\frac { \left( z+x \right) ^{2}xyz}{ \left( x+y \right) \left( x+y+z \right) \left( y+z \right) }}\geq 0. $

Correct my solution please? Write given inequality $$ \sum_{cyc} \dfrac{x+y}{2(x+y+z)} \cdot \sqrt{ z^{2}+(xy+yz+zx)} \geq \dfrac{2(xy+yz+zx)}{x+y+z} $$. Observe that function $f(t)=\sqrt {t^{2}+a}$ for fixed $a$ positive real number and $t$ being positive real, $f$ is convex.So, by assuming $xy+yz+zx=a$ is constant in above inequality, we have(by Jensen's inequality) $LHS= \sum_{cyc} \dfrac{x+y}{2(x+y+z)} \cdot f(z) \geq f \bigg(\sum_{cyc} \dfrac{z(x+y)}{2(x+y+z)} \bigg) =f \bigg(\dfrac{xy+yz+zx}{x+y+z} \bigg)=\sqrt {\dfrac{(xy+yz+zx)^{2}}{(x+y+z)^{2}}+( xy+yz+zx)}$.Let $a=xy+yz+zx$ and $b=x+y+z$.Then $\sqrt {\dfrac{(xy+yz+zx)^{2}}{(x+y+z)^{2}}+( xy+yz+zx)}=\sqrt{\dfrac{a^{2}}{b^{2}}+a}$ and so $\sqrt{\dfrac{a^{2}}{b^{2}}+a} \geq \dfrac{2(xy+yz+zx)}{x+y+z}=\dfrac{2a}{b}$ is equivalent to $b^{2} \geq 3a$ or $(x+y+z)^{2} \geq 3(xy+yz+zx)$ and latter inequality is true, since $(x+y+z)^{2}-3(xy+yz+zx)=\frac{1}{2} \cdot ((x-y)^{2}+(y-z)^{2}+(z-x)^{2}) \geq 0$

Correct solution this time. Let $x+y=2c$,$y+z=2a$ and $z+x=2b$ for positive $a,b$ and $c$. We have $x=b+c-a$,$y=c+a-b$ and $z=a+b-c$. Therefore given inequality reduces to $\sum_{cyc} 2c \sqrt{4ab} \geq 4 \big(\sum_{cyc} (a+b-c)(b+c-a) \big)$ or $\sum_{cyc} c\sqrt{ab} \geq 2ab+2bc+2ca-a^{2}-b^{2}-c^{2}$. Latter inequality is equivalent to $a^{2}+b^{2}+c^{2}-ab-bc-ca \geq ab+bc+ca-a\sqrt{bc}-b\sqrt{ca}-c\sqrt{ab}$. Note that $x^{2}+y^{2}+z^{2}-xy-yz-zx=\frac{1}{2} \big((x-y)^{2}+(y-z)^{2}+(z-x)^{2} \big)$. Using this identity for $(x,y,z)=(a,b,c) , (\sqrt{ab},\sqrt{bc},\sqrt{ca})$ , we have last inequality is equivalent to $ \sum_{cyc} (a-b)^{2} \geq \sum_{cyc} (\sqrt{ca}-\sqrt{bc})^{2} $. Now note that $ (a-b)^{2}-(\sqrt{ca}-\sqrt{bc})^{2} =(\sqrt{a}-\sqrt{b})^{2} \cdot \big((\sqrt{a}+\sqrt{b})^{2}-c \big)=(\sqrt{a}-\sqrt{b})^{2} \cdot (a+b-c+2\sqrt{ab})$ and since $a+b-c=z >0$, we have $(a-b)^{2} \geq (\sqrt{ca}-\sqrt{bc})^{2}$ so desired inequality proved

We have $RHS=\sum_{cyc}2x(y+z)$. Let $a,b,c=x+y,y+z,z+x$ then the inequality is equivalent to $$\sum_{cyc} a\sqrt{bc} \ge \sum_{cyc}2\left(\frac{a+b-c}{2}\cdot c\right)$$$$\sum_{cyc} \left(a\sqrt{bc}+c^2\right) \ge \sum_{cyc} 2ac$$ To prove the last line, let $a,b,c = p^2,q^2,r^2$. We use 4th degree Schur and then AM-GM: $$\sum_{cyc} \left(p^2qr + r^4\right) \ge \sum_{cyc} \left(p^3r + pr^3\right) \ge \sum_{cyc}2p^2r^2$$as required.

$ My $ $ solution $ $ z+x\geq 2\sqrt{zx},y+x\geq 2\sqrt{yx},x+y\geq 2\sqrt{xy} $ $ Then $ $ \sum_{cyc}(x+y)\sqrt{(z+x)(z+y)} \geq 2\sqrt{xy}\sqrt{2\sqrt{xz}2\sqrt{yz}}=4(xy+xz+yz) $

$$ \left\{ \begin{gathered} x+y= c\\ y+z=a \\ z+x=b \\ \end{gathered} \right. \Rightarrow \left\{ \begin{gathered} x=\frac{b+c-a}{2} \\ y=\frac{a+c-b}{2} \\ z=\frac{a+b-c}{2} \\ \end{gathered} \right.\Rightarrow$$$$ \Rightarrow c\sqrt{ab}+a\sqrt{bc}+b\sqrt{ac} \geq 2ab+2bc+2ac-(a^2+b^2+c^2) \Rightarrow$$$$\Rightarrow \underbrace{2ab+2bc+2ac-(a^2+b^2+c^2)-\left(c\sqrt{ab}+a\sqrt{bc}+b\sqrt{ac} \right)}_S \leq 0$$$$ 0 \geq S \geq 2ab+2bc+2ac-(a^2+b^2+c^2)-c\frac{a+b}{2}-a\frac{b+c}{2}-b\frac{a+c}{2}=$$$$ = ab+bc+ac-(a^2+b^2+c^2)=-(a^2+b^2+c^2-ab-bc-ac)=$$$$=-\frac{1}{2} \left( (a-c)^2+(c-b)^2+(b-a)^2 \right)$$

Claim:$(x+y)\sqrt{(z+x)(z+y)} \geq 2xy+yz+zx$ Proof: After we square both sides, it suffices to show $x^3y+xy^3+x^3z+y^3z \geq 2x^2y^2 +x^2yz+xy^2z$. But by AM-GM, we have $x^3y+xy^3\geq 2x^2y^2$ and $x^3z+y^3z\geq x^2yz+xy^2z\square$ So, $\sum_{cyc}(x+y)\sqrt{(z+x)(z+y)} \geq \sum_{cyc}2xy+yz+zx =4(xy+yz+zx)$ and we are done.

Solution : Let $x+y=a^2,y+z=b^2$ and $z+x=c^2$ we can write the inequality as \[\sum_{cyc}a^2bc\geq 4\sum_{cyc}xy\]we can write this as \[\sum_{cyc}a^2bc+\sum_{cyc}a^4\geq 2\sum_{cyc}a^2b^2\]we can see the LHS as $\sum_{cyc}a^2(a^2+bc)$ which is basically schur's inequality so we can write this as \[\implies \sum_{cyc}a^2(a^2+bc)\geq \sum_{cyc}a^3(b+c)\]which follows by schur's inequality so its suffince to prove that \[\sum_{sym}a^3b\geq \sum_{cyc}a^2b^2\]which is true since $(3.1)\succ(2,2)$ which is true by muirhead and we are done $\blacksquare$

me when Make the substitution $$a=x+y,b=y+z,c=z+x.$$It can then be seen that $$x=\frac{a+c-b}{2}$$etcera. Therefore, we have $$4(xy+xz+yz)=2(ab+ac+bc)-(a^2+b^2+c^2).$$Therefore, it remains to show that $$a^2+b^2+c^2+a\sqrt{bc}+b\sqrt{ca}+c\sqrt{ab}\geq 2(ab+ac+bc).$$We rewrite this as $$cyc(2,0,0)+cyc(1,1/2,1/2)\geq 2cyc(1,1,0).$$We can double the degrees (essentially the substitution $a=a_2 ^2$ etc, so it becomes $$cyc(4,0,0)+cyc(2,1,1)\geq 2cyc(2,2,0).$$By Schur with $r=2$, it follows that $$cyc(4,0,0)+cyc(2,1,1)\geq sym(3,1,0),$$and by Muirhead it then follows that $$sym(3,1,0)\geq sym(2,2,0)=2cyc(2,2,0),$$so we are done.

\begin{align*} \sum_{cyc}^{} (x + y) \sqrt { (x + z)(y + z) } &= \sum_{cyc}^{} \sqrt{ (x + y)(x + z) } \cdot \sqrt{ (x + y)(y + z) } \\ \sum_{cyc}^{} \sqrt{ (x + y)(x + z) } \cdot \sqrt{ (x + y)(y + z) } &\geq \frac {2} { \frac{1} {(x + y)(x + z)} + \frac {1} {(x + y)(y + z)}} = \sum_{cyc}^{} \frac {2(x+y)(y+z)(z + x)} {(x + z) + (y + z)} = \\ 2(x + y)(y + z)(z + x) \sum_{cyc}^{} \frac {1} {(x + z) + (y + z)} &\text{ vs } 4(xy + yz + zx) \\ (x + y)(y + z)(z + x) \sum_{cyc}^{} \frac {1} {(x + z) + (y + z)} &\text{ vs } 2(xy + yz + zx) \end{align*}Notice that: \begin{align*} &(\sum_{cyc}^{} \frac {1} {(x + z) + (y + z)})( \sum_{cyc}^{} ((x + z) + (y + z)) \geq (1 + 1 + 1)^2 \\ &4(\sum_{cyc}^{} \frac {1} {(x + z) + (y + z)})(x + y + z) \geq 9 \\ &\sum_{cyc}^{} \frac {1} {(x + z) + (y + z)} \geq \frac{9} { 4(x + y + z) } \end{align*}So we have to prove: \begin{align*} \frac{9(x + y)(y + z)(z + x)} { 4(x + y + z) } &\geq 2(xy + yz + zx) \\ 9(x + y)(y + z)(z + x) &\geq 8 (x + y + z)( xy + yz + zx ) \\ 9 (\sum_{sym}^{} x^2y + 2xyz) &\geq 8 ( \sum_{sym}^{} + 3xyz) \\ 9 \sum_{sym}^{} x^2y + 18xyz &\geq 8 \sum_{sym}^{} x^2y + 24xyz \\ \sum_{sym}^{} x^2y &\geq 6xyz \\ x^2y + x^2z + y^2z + y^2x + z^2x + z^2y &\geq 6xyz \end{align*}Which is AM-GM.

Generalization 1 Let $a_{1},a_{2},\cdots,a_{n}(n\geq 3)$ be positive reels. Then prove that $$\sum_{cyc}{\left((a_{1}+a_{2}+\cdots+a_{n-1})\sqrt[n-1]{\prod_{i=1}^{n-1}{\left(a_{n}+a_{i}\right)}}\right)}\geq 2\left(\sum_{i=k+1-n}^{n-1}{a_{i}a_{k}}\right)+(n-1)\left(\sum_{cyc}{\left(a_{1}a_{2}\cdots a_{n-1}\right)^{\dfrac{n-1}{2}}}\right)$$