Integers or Reals?

By AM-GM it suffices to prove $\left(2a^2+bc\right)\left(2b^2+ac\right)\left(2c^2+ab\right)\ge (ab+bc+ca)^3$. By homogeneity we can assume $abc=1$ hence left to prove is $\left(2a^3+1\right)\left(2b^3+1\right)\left(2c^3+1\right)\ge (ab+bc+ca)^3$ which is true by Hölder.

parmenides51 wrote: Let $a, b, c$ be positive reals. Prove that $$\sqrt{2a^2+bc}+\sqrt{2b^2+ac}+\sqrt{2c^2+ab}\ge 3 \sqrt{ab+bc+ca}$$ Let $a, b, c$ be non-negative real numbers . Prove that$$\left ( \sqrt 2+\frac{1}{2} \right )(a+b+c)\ge \sqrt{2a^2+bc}+\sqrt{2b^2+ac}+\sqrt{2c^2+ab}\ge 3 \sqrt{ab+bc+ca}$$Let $a,\,b,\,c$ are non-negative real numbers and $k \geqslant 1.$ Prove that $$\sqrt{k^2a^2+bc}+\sqrt{k^2b^2+ca}+\sqrt{k^2c^2+ab} \leqslant \left ( k+\frac{1}{2} \right )(a+b+c).$$Let $a, b, c$ be positive reals. Prove that $$a\sqrt{a^2+2bc}+b\sqrt{b^2+2ac}+c\sqrt{c^2+2ba} \ge \sqrt{3}(ab+bc+ca) $$

A different proof. Note that taking squares, and using the trivial inequality $a^2+b^2+c^2\geqslant ab+bc+ca$, it boils down proving $$ \sum_{{\rm cyc}}\sqrt{(2a^2+bc)(2b^2+ac)}\geqslant 3(ab+bc+ca). $$Now, observe that using Cauchy-Schwarz, $$ \sum_{{\rm cyc}}\sqrt{(a^2+a^2+bc)(ac+b^2+b^2)}\geqslant \sum_{{\rm cyc}}(a^{3/2}c^{1/2}+ab+b^{3/2}c^{1/2}). $$With this move, it boils down proving $$ \sum_{{\rm cyc}}(a^{3/2}c^{1/2}+b^{3/2}c^{1/2})\geqslant 2(ab+bc+ca). $$Letting $a=m^2,b=n^2,c=k^2$, this is equivalent to proving $\sum_{{\rm cyc}}m^3 n\geqslant 2(m^2n^2+n^2k^2+k^2m^2)$. Since $(3,1,0)$ majorizes $(2,2,0)$, Muirhead inequality then finishes the rest.

sqing wrote: Let $a, b, c$ be positive reals. Prove that $$a\sqrt{a^2+2bc}+b\sqrt{b^2+2ac}+c\sqrt{c^2+2ba} \ge \sqrt{3}(ab+bc+ca) $$ By Holder: \[\text{LHS}^2 . \sum\limits_{cyc} a(a^2+2bc)^2 \geqq \Big[\sum\limits_{cyc} a(a^2+2bc)\Big]^3 \]Hence$,$ we need to prove: \[ \Big[\sum\limits_{cyc} a(a^2+2bc)\Big]^3 \geqq 3(ab+bc+ca)^2 . \sum\limits_{cyc} a(a^2+2bc)^2\]The last one is true$,$ who have a nice SOS for it$?$ My SOS is very very ugly

This is stronger: Let $a, b, c$ be positive reals. Prove that $$\sqrt{2a^2+bc}+\sqrt{2b^2+ac}+\sqrt{2c^2+ab}\ge \sqrt{2(a^2+b^2+c^2)+7(ab+bc+ca)}$$

ant_ wrote: This is stronger: Let $a, b, c$ be positive reals. Prove that $$\sqrt{2a^2+bc}+\sqrt{2b^2+ac}+\sqrt{2c^2+ab}\ge \sqrt{2(a^2+b^2+c^2)+7(ab+bc+ca)}$$ This can be done in almost the same way as in #5, the difference is you square both sides first and then use AM-GM, not the other way around. The rest is identical. sqing wrote: Let $a, b, c$ be positive reals. Prove that $$a\sqrt{a^2+2bc}+b\sqrt{b^2+2ac}+c\sqrt{c^2+2ba} \ge \sqrt{3}(ab+bc+ca) $$ This is true by Chebyshev. If WLOG $a\ge b\ge c$ then $$LHS\ge\frac{1}{3}\cdot\sum a\cdot\sum\sqrt{2a^2+bc}\ge\frac{1}{3}\cdot\sqrt{3(ab+bc+ca)}\cdot 3\sqrt{ab+bc+ca}=RHS$$

parmenides51 wrote: Let $a, b, c$ be positive reals. Prove that $$\sqrt{2a^2+bc}+\sqrt{2b^2+ac}+\sqrt{2c^2+ab}\ge 3 \sqrt{ab+bc+ca}$$ Let $ a,b,c\ge 0$ and $ab+ac+bc=2$.Prove that $$\sqrt{a+ab}+\sqrt{b+bc}+\sqrt{c+ca}\ge3$$

parmenides51 wrote: Let $a, b, c$ be positive reals. Prove that $\sqrt{2a^2+bc}+\sqrt{2b^2+ac}+\sqrt{2c^2+ab}\ge 3 \sqrt{ab+bc+ca}$ We have by minowski and AM-GM $$\sum\sqrt{2a^2+bc}= \sqrt{bc+a^2+a^2}+\sqrt{b^2+ac+b^2}+\sqrt{c^2+c^2+ab}\geq \sqrt{\sum (\sqrt{bc}+b+c)^2}\geq 3\sqrt{bc}$$

After AM-GM it suffices to show $(2a^2 + bc)(2b^2 + ac)(2c^2 + ab) \geq (ab+bc+ac)^3$. After expanding, this is equivalent to $9a^2b^2c^2 + 4\sum{a^3b^3} + 2\sum{a^4bc} \geq \sum{a^3b^3} + 3\sum{a^3b^2c} + 3\sum{b^3a^2c} + 6a^2b^2c^2 \iff 3a^2b^2c^2 + 3\sum{a^3b^3} + 2\sum{a^4bc} \geq 3\sum{abc(a^2b + b^2a)}$ but by AM-GM, we know that $a^2b^2c^2 + a^3b^3 + a^4bc \geq 3a^3b^2c$, so after adding all of these, it suffices to show $\sum{a^3b^3} \geq 3a^2b^2c^2$ but this is true by AM-GM.

First of all notice that $\sum_{cyc}\sqrt{2a^2+bc}\overset{\text{AM-GM}}{\ge}3\sqrt[3]{\sqrt{\prod_{cyc}2a^2+bc}}=3\sqrt[3]{\sqrt{9a^2b^2c^2+4\sum_{cyc}a^3b^3+2abc\sum_{cyc}a^3}}$ Thus the inequality transforms into $$\left(9a^2b^2c^2+4\sum_{cyc}a^3b^3+2abc\sum_{cyc}a^3\right)^{1/6}\ge\left(\sum_{cyc}ab\right)^{1/2}\Longleftrightarrow9a^2b^2c^2+4\sum_{cyc}a^3b^3+2abc\sum_{cyc}a^3\ge\left(\sum_{cyc}ab\right)^3=\sum_{cyc}a^3b^3+3abc\sum_{cyc}a^2b+3abc\sum_{cyc}ab^2+9a^2b^2c^2$$This furthermore boils down to $3a^2b^2c^2+3\sum_{cyc}a^3b^3+2abc\sum_{cyc}a^3\ge3abc\sum_{cyc}a^2b+3abc\sum_{cyc}ab^2$ However notice that $a^2b^2c^2+a^3b^3+a^4bc\overset{\text{AM-GM}}{\ge}3a^3b^2c$ and by summing cyclically the inequality becomes $2\sum_{cyc}a^3b^3+abc\sum_{cyc}a^3\ge3\sum_{cyc}b^3a^2c$ Furthermore $\sum_{cyc}a^3b^3\ge\sum_{cyc}b^3a^2c$ by $\text{Muirhead}$ since $(3,3,0)\succ(3,2,1)$ thus all that we have left to prove is $abc\sum_{cyc}a^3\ge abc\sum_{cyc}b^2a\Longleftrightarrow\sum_{cyc}a^3\ge\sum_{cyc}b^2a$ However this is also true by $\text{Muirhead}$ since $(3,0,0)\succ(2,1,0)$

Let $a, b, c$ be positive reals. Prove that $$\sqrt{a^2+2bc}+\sqrt{b^2+2ca}+\sqrt{c^2+2ab}\ge 3 \sqrt{ab+bc+ca}$$h h

parmenides51 wrote: Let $a, b, c$ be positive reals. Prove that $$\sqrt{2a^2+bc}+\sqrt{2b^2+ac}+\sqrt{2c^2+ab}\ge 3 \sqrt{ab+bc+ca}$$ Let $a,b,c$ be nonnegative real numbers.If $ 0 \leq k \leq \dfrac{5+2\sqrt{6}}{2} $, Prove that \[ \sqrt{a^2+kbc}+\sqrt{b^2+kca}+\sqrt{c^2+kab} \geq \sqrt{3(k+1)(ab+bc+ca)}.\]2000 KJMO: Let $ a,b,c>0 $ and $ abc\geq a+b+c. $ Prove that $$\sqrt{2a^2+bc}+\sqrt{2b^2+ac}+\sqrt{2c^2+ab}\geq 9$$

@grupyorum Hey, can you please explain how have you used Muirhead inequality? It's just I'm quite confused how to deal with constants like "2" in our case. "Since (3, 1, 0) majorizes (2, 2, 1) Muirhead inequality then finishes the rest.", why not like that?

Pashtet671 wrote: @grupyorum Hey, can you please explain how have you used Muirhead inequality? It's just I'm quite confused how to deal with constants like "2" in our case. "Since (3, 1, 0) majorizes (2, 2, 1) Muirhead inequality then finishes the rest.", why not like that? It was a typo. It means: $$\sum_{cyc}(m^3n+m^3k)\geq2\sum_{cyc}m^2n^2.$$Like your typo: Pashtet671 wrote: $(3, 1, 0)$ majorizes $(2, 2, 1)$

Or symmetric $\sum\limits_{sym}{m^3n}$ sum, for the Muirhead.

Victoria_Discalceata1 wrote: ant_ wrote: This is stronger: Let $a, b, c$ be positive reals. Prove that $$\sqrt{2a^2+bc}+\sqrt{2b^2+ac}+\sqrt{2c^2+ab}\ge \sqrt{2(a^2+b^2+c^2)+7(ab+bc+ca)}$$ This can be done in almost the same way as in #5, the difference is you square both sides first and then use AM-GM, not the other way around. The rest is identical. sqing wrote: Let $a, b, c$ be positive reals. Prove that $$a\sqrt{a^2+2bc}+b\sqrt{b^2+2ac}+c\sqrt{c^2+2ba} \ge \sqrt{3}(ab+bc+ca) $$ This is true by Chebyshev. If WLOG $a\ge b\ge c$ then $$LHS\ge\frac{1}{3}\cdot\sum a\cdot\sum\sqrt{2a^2+bc}\ge\frac{1}{3}\cdot\sqrt{3(ab+bc+ca)}\cdot 3\sqrt{ab+bc+ca}=RHS$$ I think your Chebyshev is wrong because, $\left\{a,b,c\right\}$ and $\left\{\sqrt{2a^2+bc},\sqrt{2b^2+ac},\sqrt{2c^2+ab}\right\}$ aren't similarly ordered