3 var inquality

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sqing

#1 h Apr 19, 2025, 8:32 AM

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Let $a,b,c$ be reals such that $a+b+c=0$ and $abc\geq \frac{1}{\sqrt{2}} .$ Prove that
$a^2+b^2+c^2\geq 3$ Let $a,b,c$ be reals such that $a+2b+c=0$ and $abc\geq \frac{1}{\sqrt{2}} .$ Prove that
$a^2+b^2+c^2\geq \frac{3}{ \sqrt[3]{2}}$ $a^2+2b^2+c^2\geq 2\sqrt[3]{4}$

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#2 h Apr 19, 2025, 11:38 AM

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sqing wrote:

Let $a,b,c$ be reals such that $a+b+c=0$ and $abc\geq \frac{1}{\sqrt{2}} .$ Prove that
$a^2+b^2+c^2\geq 3$ Let $a,b,c$ be reals such that $a+2b+c=0$ and $abc\geq \frac{1}{\sqrt{2}} .$ Prove that
$a^2+b^2+c^2\geq \frac{3}{ \sqrt[3]{2}}$ $a^2+2b^2+c^2\geq 2\sqrt[3]{4}$

$a+b+c=0 \iff b=-(a+c)$ $abc\geq \frac{1}{\sqrt{2}} \iff ac(a+c) \leq -\frac{1}{\sqrt 2}$ Let $u=a+c$ , $v=ac$
if $u>0$ then we have $v<0 \iff a>0>c$
if $u<0$ then we have $v>0 \iff c<a<0$
==> $a^2+b^2+c^2\geq 3 \iff 2(u^2-v) \geq 3$ equality cases : $(a,b,c)=(-\frac{1}{\sqrt 2},-\frac{1}{\sqrt 2},\sqrt 2)$ and permutations.

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#3 h Apr 19, 2025, 11:40 AM

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the second one $(a,b,c)=(-2^{-\frac{1}{6}},2^{-\frac{1}{6}},-2^{-\frac{1}{6}})$
the third one $(a,b,c)=(-2^{-\frac{1}{6}},2^{-\frac{1}{6}},-2^{-\frac{1}{6}})$

This post has been edited 1 time. Last edited by SunnyEvan, Apr 19, 2025, 12:22 PM

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sqing

#4 h Apr 19, 2025, 12:19 PM

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SunnyEvan wrote:

the second one $(a,b,c)=(-2^{\frac{1}{6}},2^{\frac{1}{6}},-2^{\frac{1}{6}})$
the third one $(a,b,c)=(-2^{\frac{1}{6}},2^{\frac{1}{6}},-2^{\frac{1}{6}})$

the second one $(a,b,c)=(-2^{-\frac{1}{6}},2^{-\frac{1}{6}},-2^{-\frac{1}{6}})$
the third one $(a,b,c)=(-2^{-\frac{1}{6}},2^{-\frac{1}{6}},-2^{-\frac{1}{6}})$

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sqing

#5 h Apr 19, 2025, 12:21 PM

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Nice.Thank SunnyEvan.

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sqing

#6 h Apr 20, 2025, 2:45 AM

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sqing wrote:

Let $a,b,c$ be reals such that $a+4b+c=0$ and $abc\geq \frac{1}{\sqrt{2}} .$ Prove that $a^2+b^2+c^2\geq \frac{9}{2\sqrt[3]{4}}$

Solution of DAVROS:
$a,b,c\ne0$ and only one can be positive so check cases : $b<0, a>0, c<0$ and $b>0, a<0, c<0$

$b<0, a>0, c<0, ac<0, a+c=-4b, ac \le\frac1{b\sqrt2}$

$\textbf{LHS} = (a+c)^2-2ac+b^2 \ge 17b^2-\frac{\sqrt2}b = 17(-b)^2+2\cdot \frac1{(-b)\sqrt2} \ge 3\sqrt[3]{\frac{17}2}>6$

$b>0, a<0, c<0, ac>0, a +c=-4b, ac \ge \frac1{b\sqrt2}$

$(a+c)^2\ge 4ac \implies 16b^2\ge \frac4{b\sqrt2} \implies b \ge \frac1{\sqrt[6]{32}}$

$\textbf{LHS} \ge \frac{(a+c)^2}2 +b^2 = 9b^2 \ge \frac9{2^{\frac53}}$ at $b=\frac1{\sqrt[6]{32}}, a=c=-2b\quad\blacksquare$

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