The following inequality is also true

This will be a good result but a solution please?

Schur inequality

I agree with you on this way, but I can't realised how to solved it. Because RHS isn't symetric. For me, we have to do some changes on condition to get a simple experission. Can anyone help me for solve this question? At least can anyone give a useful hint for arrange the condition? Thanks.

mberke wrote: Let $a,b,c$ be non-negative real numbers such that $a^2+b^2+c^2 \le 3$ then prove that; $$(a+b+c)(a+b+c-abc)\ge2(a^2b+b^2c+c^2a)$$ It is very amazing for an easy inequality in Turkey TST. Note that $a+b+c\leq \sqrt{3(a^2+b^2+c^2)}\leq 3,$ and $a+b+c \geq \dfrac{(a+b+c)(a^2+b^2+c^2)}{3}\geq 3abc$, we get $$ abc(a+b+c)=\dfrac{2abc(a+b+c)}{3}+\dfrac{abc(a+b+c)}{3} \leq 2abc+\dfrac{(a+b+c)^2}{9}.$$It suffices to show that $$ (a+b+c)^2-2abc-\dfrac{(a+b+c)^2}{9} \geq 2(a^2b+b^2c+c^2a),$$which is equivalent to $$ \dfrac{4(a+b+c)^2}{9} \geq a^2b+b^2c+c^2a+abc.$$Now, since $ \dfrac{4(a+b+c)^2}{9} \geq \dfrac{4(a+b+c)^3}{27},$ we only need to prove that $$ \dfrac{4(a+b+c)^3}{27} \geq a^2b+b^2c+c^2a+abc,$$which is known Vasc's inequality.

luofangxiang wrote: The following inequality is also true It follows from Vasc's inequality $$ 4(a+b+c)^3-27(a^2b+b^2c+c^2a)\geq 27abc\geq 3abc(a+b+c)^2.$$

From the condition, we get that $ a+b+c \le 3 $. Note that it is enough to prove the inequality when $ a+b+c =3 $, so the given inequality is equivalent with $ (a+b+c)^3 -9abc \ge 6(a^2b+b^2c+c^2a) $, or $ a^3+b^3+c^3+3ab^2+3bc^2+3ca^2 \ge 3abc+3a^2b+3b^2c+3c^2a $ Since $ ab^2+bc^2+ca^2 \ge 3abc $, we are left to prove that $ a^3+b^3+c^3+2ab^2+2bc^2+2ca^2 \ge 3a^2b+3b^2c+3c^2a $ To prove this, we can assume that $ c \le a, c\le b $ so we can express $ a=x+c, b=y+c, x,y\ge 0 $ and after substituting them we get a simple inequality which can be easily proved.

andrejilievski wrote: and after substituting them we get a simple inequality which can be easily proved. A lot of calculations is needed here...

Let $a,b,c$ be non-negative real numbers such that $a^2+b^2+c^2 \le 3$ then prove that; $$\sqrt{a+b+c}(a+b+c-abc)\ge2\sqrt{a^2b+b^2c+c^2a}$$

sqing wrote: Let $a,b,c$ be non-negative real numbers such that $a^2+b^2+c^2 \le 3$ then prove that; $$\sqrt{a+b+c}(a+b+c-abc)\ge2\sqrt{a^2b+b^2c+c^2a}$$ Let $a=kx$, $b=ky$ and $c=kz$, where $k>0$ and $x^2+y^2+z^2=3$. Hence, $k\leq1$ and we need to prove that $\sqrt{x+y+z}(x+y+z-k^2xyz)\geq2\sqrt{x^2y+y^2z+z^2x}$. But $\sqrt{x+y+z}(x+y+z-k^2xyz)\geq\sqrt{x+y+z}(x+y+z-xyz)$. Thus, it remains to prove that $\sqrt{x+y+z}(x+y+z-xyz)\geq2\sqrt{x^2y+y^2z+z^2x}$, which is $\sqrt{x+y+z}((x^2+y^2+z^2)(x+y+z)-3xyz)\geq2(x^2+y^2+z^2)\sqrt{x^2y+y^2z+z^2x}$ or $\sum_{cyc}(x^7-x^6y+3x^6z+5x^5y^2+x^5z^2-x^4y^3+7x^4z^3-7x^4y^2z-11x^4z^2y+3x^3y^2z^2)\geq0$, which is obvious.

sqing wrote: Let $a,b,c$ be non-negative real numbers such that $a^2+b^2+c^2 \le 3$ then prove that; $$\sqrt{a+b+c}(a+b+c-abc)\ge2\sqrt{a^2b+b^2c+c^2a}$$

Sorry, How to prove the last inequality?

Schur inequality

Let $a,b,c$ be non-negative real numbers such that $ab+bc+ca \le 3$ then prove that; $$(a+b+c-1)(a^3+b^3+c^3-abc)\ge2(a^3b+b^3c+c^3a-1)$$

sqing wrote: Let $a,b,c$ be non-negative real numbers such that $ab+bc+ca \le 3$ then prove that; $$(a+b+c)(a^3+b^3+c^3-abc)\ge2(a^3b+b^3c+c^3a)$$ The following inequality is also true

luofangxiang wrote: The following inequality is also true We have \[LHS-RHS=\frac{1}{2}\sum{(a^2-b^2-ab+bc)^2}\ge{0}\]

$$(a+b+c)(a+b+c-abc)\ge2(a^2b+b^2c+c^2a)$$this inequality also hold when $a^2+b^2+c^2+abc \le 4$

mberke wrote: Let $a,b,c$ be non-negative real numbers such that $a^2+b^2+c^2 \le 3$ then prove that; $$(a+b+c)(a+b+c-abc)\ge2(a^2b+b^2c+c^2a)$$ Hey #mberke can you send me all problems of TURKEY İMO TST 2016?????

You can find it these questions on http://www.artofproblemsolving.com/community/c256440_2016_turkey_team_selection_test

quykhtn-qa1 wrote: luofangxiang wrote: The following inequality is also true It follows from Vasc's inequality $$ 4(a+b+c)^3-27(a^2b+b^2c+c^2a)\geq 27abc\geq 3abc(a+b+c)^2.$$ sorry... what is vasc's inequality, and how can we prove it??

We need to prove following inequality: $$ (a+b+c)^2\geq 2(a^2b+b^2c+c^2a)+abc(a+b+c).$$From $a^2+b^2+c^2 \le 3$ we easly conclude that $a+b+c\le 3$ Now, after multiplying the inequality by $3$ we need to prove: $$ 3(a+b+c)^2\geq 6(a^2b+b^2c+c^2a)+3abc(a+b+c).$$But we have that: $$ 3(a+b+c)^2\geq(a^2+b^2+c^2)(a+b+c)^2=((a^2+b^2+c^2))^2+2(a^2+b^2+c^2)(ab+bc+ca)$$and $$6(a^2b+b^2c+c^2a)\leq2(a+b+c)(a^2b+b^2c+c^2a)$$So the inequality is equivalent to: $${(a^2+b^2+c^2)}^2+2(a^2+b^2+c^2)(ab+bc+ca)\geq2(a+b+c)(a^2b+b^2c+c^2a)+3abc(a+b+c)$$It is not hard to tee that $$2(a^2+b^2+c^2)(ab+bc+ca)-2(a+b+c)(a^2b+b^2c+c^2a)=2(ab^3+bc^3+ca^3-a^2b^2-b^2c^2-c^2a^2)$$so we need to prove that: $${(a^2+b^2+c^2)}^2-3abc(a+b+c)+2(ab^3+bc^3+ca^3-a^2b^2-b^2c^2-c^2a^2)\geq0$$or $$a^4+b^4+c^4+2(ab^3+bc^3+ca^3)-3abc(a+b+c)\geq0$$By $$ AM\geq GM $$inequality we have: $$ a^3c+a^3c+a^3c+a^3c+b^3a+b^3a+c^3a\geq7\sqrt[7]{a^14 b^7c^7}=7a^2bc$$If we do the same process, we will have that following inequality is true: $$ab^3+bc^3+ca^3\geq abc(a+b+c)$$so finally we need to prove that: $$a^4+b^4+c^4\geq abc(a+b+c)$$This is true,since: $$a^4+b^4+c^4\geq\dfrac{(a^2+b^2+c^2)^2}{3}\geq\dfrac{(ab+bc+ca)^2}{3}\geq abc(a+b+c)$$because $$ab+bc+ca\geq\sqrt{3abc(a+b+c}$$Equality holds for $$a=b=c=1.$$

Bergo1305 wrote: $$6(a^2b+b^2c+c^2a)\leq2(a+b+c)(a^2b+b^2c+c^2a)$$ This is not true, since $3\ge a+b+c$, as you also mentioned.

Solved with Guendabiaani

Not my solution, just thought it was nice, and it's pretty distinct to the other ones presented here. $$(a-ab)^2+(b-bc)^2+(c-ca)^2 \geq{0}$$Therefore, it is enough to prove that: $$2(ab+bc+ca)\geq{}a^2b^2+b^2c^2+c^2a^2+abc(a+b+c)$$Which is true after homogenising with $a^2+b^2+c^2\leq{3}$, and using AM-GM or Muirhead.

Note that \begin{align*} a^2+b^2+c^2+\frac{2}{3}(ab+bc+ca)(a^2+b^2+c^2)&\geq 2(a^2b+b^2c+c^2a)+abc(a+b+c)\Longleftrightarrow\\ 3(a^2+b^2+c^2)+2(ab+bc+ca)(a^2+b^2+c^2)&\geq 3abc(a+b+c)+6(a^2b+b^2c+c^2a)\Longleftrightarrow\\ 3(a^2+b^2+c^2)+2(a^3b+ab^3+a^3c+ac^3+b^3c+bc^3)&\geq abc(a+b+c)+6(a^2b+b^2c+c^2a). \end{align*}By Muirhead-type AM-GM, $ab^3+bc^3+ca^3\geq abc(a+b+c)$. Also, observe that $$\frac{2a^3b+ab^3+2.5a^2+0.5b^2}{6}\geq a^2b$$by AM-GM, we do the same for others and we get the rest of the inequality nuked, we are done.

mberke wrote: Let $a,b,c$ be non-negative real numbers such that $a^2+b^2+c^2 \le 3$ then prove that; $$(a+b+c)(a+b+c-abc)\ge2(a^2b+b^2c+c^2a)$$

sqing wrote: Let $a,b,c$ be non-negative real numbers such that $a^2+b^2+c^2 \le 3$ then prove that; $$\sqrt{a+b+c}(a+b+c-abc)\ge2\sqrt{a^2b+b^2c+c^2a}$$

$i)$ If $\sum{a^2bc}\geq \sum{ab}$ Let $a+b+c=3u,ab+bc+ca=3v^2,abc=w^3$ We have $uw^3\geq v^2$ and $1\geq v$. \[9u^2-3uw^3\overset{?}{\geq}9u^2-6v^2+9v^2-6uw^3\geq 9u^2-6v^2+9v^4-6uw^3=\sum{a^2}+\sum{a^2b^2}\geq 2(a^2b+b^2c+c^2a)\]This is true since $9u^2-3uw^3\geq 9u^2-6v^2+9v^2-6uw^3\iff uw^3\geq v^2.\square$ $ii)$ If $\sum{ab}\geq \sum{a^2bc}$ \[\sum{a^2}+2\sum{ab}-\sum{a^2bc}\geq \sum{a^2}+\sum{ab}\overset{?}{\geq}2(a^2b+b^2c+c^2a)\]By Vasc inequality, we have $3\sum{a^2}\geq (\sum{a^2})^2\geq 3(a^3b+b^3c+c^3a)\implies \sum{a^2}\geq (a^3b+b^3c+c^3a)$ Hence \[\sum{a^2}+\sum{ab}\geq \sum{a^3b}+\sum{ab}=\sum{ab(a^2+1)}\geq 2\sum{a^2b}\]As desired.$\blacksquare$