The inequality $9(x+y)(y+z)(z+x)\geq 8(xy+yz+zx)(x+y+z)$ (1) it seems to be lovely Using (1) and after Cauchy We have $\frac{27}{4}(x+y)(y+z)(z+x)\geq 6(x+y+z)\geq (\sqrt{x+y}+\sqrt{y+z}+\sqrt{z+x})^2$ Now by using AM-GM and then (1) we have $(\sqrt{x+y}+\sqrt{y+z}+\sqrt{z+x})^2\geq 9\sqrt[3]{(x+y)(y+z)(z+x)}\geq 9\sqrt[3]{\frac{8(x+y+z)}{9}}\geq 9 \sqrt[3]{\frac{8 \sqrt{3(xy+yz+zx)}}{9}}\geq 6\sqrt{3}$ and we are done

The inequality is very easy if we substitute $x=tan\frac{A}{2}, y=tan\frac{B}{2},z=tan\frac{c}{2}$ where $A,B,C$ are angles of the triangle. Then we get easy inequalities(very typical) that are followed by Jensen.

Looks so familiar. I guess I posted my solution too early in the wrong thread.

substitute by cotga, cotgb, cotgc so that a+b+c=180° in a acute triangle and use jensen

Can someone explain the Jensen solution fully? I tried the cotangent trick but couldn't get it to work.

Valentin Vornicu wrote: If $x,y,z$ are positive real numbers and $xy+yz+zx=1$ prove that \[ \frac{27}{4} (x+y)(y+z)(z+x) \geq ( \sqrt{x+y} +\sqrt{ y+z} + \sqrt{z+x} )^2 \geq 6 \sqrt 3. \] But you do this stronger: If $x,y,z$ are positive real numbers and $xy+yz+zx=1$ prove that \[\left( 2\,\sqrt {2}+3 \right) \left( x+y \right) \left( y+z \right) \left( z+x \right) +8\, \left( {\frac {15}{4}}-2\,\sqrt {2} \right) xyz\geq \left( \sqrt {x+y}+\sqrt {y+z}+\sqrt {z+x} \right) ^{2}\]

xzlbq wrote: Valentin Vornicu wrote: If $x,y,z$ are positive real numbers and $xy+yz+zx=1$ prove that \[ \frac{27}{4} (x+y)(y+z)(z+x) \geq ( \sqrt{x+y} +\sqrt{ y+z} + \sqrt{z+x} )^2 \geq 6 \sqrt 3. \] But you do this stronger: If $x,y,z$ are positive real numbers and $xy+yz+zx=1$ prove that \[\left( 2\,\sqrt {2}+3 \right) \left( x+y \right) \left( y+z \right) \left( z+x \right) +8\, \left( {\frac {15}{4}}-2\,\sqrt {2} \right) xyz\geq \left( \sqrt {x+y}+\sqrt {y+z}+\sqrt {z+x} \right) ^{2}\] By C-S,need to prove \[ \left( 2\,\sqrt {2}+3 \right) \left( x+y \right) \left( y+z \right) \left( z+x \right) +8\, \left( {\frac {15}{4}}-2\,\sqrt {2} \right) xyz\geq 6\,x+6\,y+6\,z\] <=> \[x{z}^{2}+y{z}^{2}+{x}^{2}y+x{y}^{2}+{x}^{2}z-6\,xyz+{y}^{2}z\geq 0\]

Valentin Vornicu wrote: If $x,y,z$ are positive real numbers and $xy+yz+zx=1$ prove that \[ \frac{27}{4} (x+y)(y+z)(z+x) \geq ( \sqrt{x+y} +\sqrt{ y+z} + \sqrt{z+x} )^2 \geq 6 \sqrt 3. \] Stronger is: If $x,y,z$ are positive real numbers and $xy+yz+zx=1$ prove that \[ \left( \sqrt {x+y}+\sqrt {y+z}+\sqrt {z+x} \right) ^{2}\geq 6\,\sqrt { \left( x+y+z \right) \sqrt {3}}\]

xzlbq wrote: Valentin Vornicu wrote: If $x,y,z$ are positive real numbers and $xy+yz+zx=1$ prove that \[ \frac{27}{4} (x+y)(y+z)(z+x) \geq ( \sqrt{x+y} +\sqrt{ y+z} + \sqrt{z+x} )^2 \geq 6 \sqrt 3. \] Stronger is: If $x,y,z$ are positive real numbers and $xy+yz+zx=1$ prove that \[ \left( \sqrt {x+y}+\sqrt {y+z}+\sqrt {z+x} \right) ^{2}\geq 6\,\sqrt { \left( x+y+z \right) \sqrt {3}}\] $W=2z+x+y$ By Holder,need to prove \[{\frac { \left( {\it \sum} \left( \left( x+y \right) \left( 2\,z+x+y \right) \right) \right) ^{3}}{{\it \sum} \left( \left( x+y \right) ^{2} \left( 2\,z+x+y \right) ^{3} \right) }}\geq 6\,\sqrt { \left( x+y+z \right) \sqrt {3}}\]<=> ${\frac { \left( 8\, \left( x+y+z \right) ^{6}+24\, \left( x+y+z \right) ^{4}+24\, \left( x+y+z \right) ^{2}+8 \right) ^{2}}{ \left( 2 \, \left( x+y+z \right) ^{5}+ \left( x+y+z \right) ^{3}+3\, \left( x+y +z \right) ^{2}xyz+7\,x+7\,y+7\,z-5\,xyz \right) ^{2}}}\geq 36\, \left( x+ y+z \right) \sqrt {3} $ very hard!

Let's first show the left inequality. After homogenizing, we want to show that: $$27(x+y)(y+z)(z+x) \ge 4(xy+yz+zx)(\sqrt{x+y} + \sqrt{y+z}+\sqrt{z+x})^2.$$ First of all, notice that $27(x+y)(y+z)(z+x) \ge 24(x+y+z)(xy+yz+zx)$ by AM-GM. Therefore, it suffices to show that $$24(x+y+z) \ge (\sqrt{x+y}+\sqrt{y+z}+\sqrt{z+x})^2.$$ However, this is easy, because $(\sqrt{x+y}+\sqrt{y+z}+\sqrt{z+x})^2 \le 3(x+y+y+z+z+x) = 6(x + y + z).$ Let's now show the second inequality. We want to show that $$(\sqrt{x+y}+ \sqrt{y+z} + \sqrt{z+x})^2 \ge 6 \sqrt3,$$ whenever $xy+yz+zx = 1$ and $x, y, z$ are nonnegative reals. If any of $x, y, z$ are bigger than $100$, then we are clearly done. Else, let us restrict $x, y, z$ to the interval $[0, 100].$ Let's prove this with Lagrange Multipliers. The derivative of the LHS with respect to $x$ is $$2(\sqrt{x+y}+ \sqrt{y+z} + \sqrt{z+x}) (\frac{1}{2 \sqrt{x+y}}+ \frac{1}{2\sqrt{x+z}}).$$ The derivative of the condition w.r.t $x$ is $y+z$. Therefore, we have that: $$\frac{\frac{1}{\sqrt{x+y}} + \frac{1}{\sqrt{x+z}}}{y+z}$$ is cyclically constant. After multiplying by $\sqrt{(x+y)(y+z)(z+x)}$, we get that: $$\frac{\sqrt{x+y} + \sqrt{x+z}}{\sqrt{y+z}}$$ is cyclically constant. This is easily seen to imply that $x = y =z.$ Hence, the minimum is achieved either when $xyz = 0$, $\max(x, y, z) = 100$, or $x = y = z = \frac{1}{\sqrt3}.$ It's easy to check that all of these cases are wins. $\square$

Surprisingly simple. Let $p = x + y + z , r = xyz$. Note that $p^2 \geq 3(xy + yz + zx) = 3 \implies p \geq \sqrt{3}$ and $r \leq \frac{1}{3\sqrt{3}}$ by AM-GM. First we show the left part. It is equivalent to $$\frac{27}{4}(p - r) \geq 2p + 2\sum_{\text{cyc}} \sqrt{x^2 + 1}$$By C-S, $\sum_{\text{cyc}} \sqrt{x^2 + 1} \leq \sqrt{3(x^2 + y^2 + z^2 + 3)} = \sqrt{3(p^2 + 1)}$. So it suffices to show $\frac{19}{4}p - 2\sqrt{3(p^2 + 1)} \geq \frac{27}{4}r$. Note that the LHS is increasing in $p$ and so $\frac{19}{4}p - 2\sqrt{3(p^2 + 1)} \geq \frac{3\sqrt{3}}{4} \geq \frac{27}{4}r$. Now let's show the right part. It is equivalent to $$2p + 2\sum_{\text{cyc}} \sqrt{x^2 + 1} \geq 6\sqrt{3}$$. Since $\sqrt{x^2 + 1}$ is convex, by Jensen we get $2\sum_{\text{cyc}} \sqrt{x^2 + 1} \geq 4\sqrt{3}$. And since $2p \geq 2\sqrt{3}$, we get the result. Done.