Let $a, b, c$ be positive reals. Prove that $\sqrt{2a^2+bc}+\sqrt{2b^2+ac}+\sqrt{2c^2+ab}\ge 3 \sqrt{ab+bc+ca}$
Problem
Source: Ukraine TST 2012 p1
Tags: inequalities, algebra
30.04.2020 02:02
Integers or Reals?
30.04.2020 02:05
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.
30.04.2020 02:35
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 $$a\sqrt{a^2+2bc}+b\sqrt{b^2+2ac}+c\sqrt{c^2+2ba} \ge \sqrt{3}(ab+bc+ca) $$
30.04.2020 03:45
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.
02.05.2020 06:20
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
06.08.2020 08:14
This is stronger: Let $a, b, c$ be positive reals. Prove that $$\sqrt{2a^2+bc}+\sqrt{2b^2+ac}+\sqrt{2c^2+ab}\ge \sqrt{2(a^2+b^2+c^2)+7(ab+bc+ca)}$$
06.08.2020 09:47
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 $$LHS\ge\frac{1}{3}\cdot\sum a\cdot\sum\sqrt{2a^2+bc}\ge\frac{1}{3}\cdot\sqrt{3(ab+bc+ca)}\cdot 3\sqrt{ab+bc+ca}=RHS$$
27.10.2020 09:43
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 $$\sqrt{a+ab}+\sqrt{b+bc}+\sqrt{c+ca}\ge3$$
26.02.2021 19:57
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 $$\sum\sqrt{2a^2+bc}= \sqrt{bc+a^2+a^2}+\sqrt{b^2+ac+b^2}+\sqrt{c^2+c^2+ab}\geq \sqrt{\sum (\sqrt{bc}+b+c)^2}\geq 3\sqrt{bc}$$
02.04.2021 10:07
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.
09.08.2023 00:46
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 $(3,0,0)\succ(2,1,0)$
09.08.2023 05:14
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
19.02.2024 22:18
30.06.2024 04:24
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 $$\sqrt{2a^2+bc}+\sqrt{2b^2+ac}+\sqrt{2c^2+ab}\geq 9$$
31.07.2024 10:19
@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?
31.07.2024 13:28
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 $(2, 2, 1)$
03.08.2024 10:45
Or symmetric $\sum\limits_{sym}{m^3n}$ sum, for the Muirhead.
15.01.2025 08:20
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