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Let $a$, $b$ and $c$ be non-negative reals such that $ab+bc+ca=6$.

Prove that:

$$\frac{(a+b)^{3}}{\sqrt[3]{2(a+b)(a^{2}+b^{2})}}+\frac{(b+c)^{3}}{\sqrt[3]{2(b+c)(b^{2}+c^{2})}}+\frac{(c+a)^{3}}{\sqrt[3]{2(c+a)(c^{2}+a^{2})}}\ge{12}$$

Oshawott
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user136728
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3 Answers3

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Applying AM-GM on $2ab$ and $a^2+b^2$ we find $2ab(a^2+b^2) \leq \left(\frac{(a+b)^2}{2} \right)^2$ or $a^2+b^2 \leq \frac{(a+b)^4}{8ab}$. Plugging this in, we find $$ \frac{(a+b)^3}{\sqrt[3]{2(a+b)(a^2+b^2)}} \geq \frac{(a+b)^3}{\sqrt[3]{2(a+b)\frac{(a+b)^4}{8ab}}} = \frac{(a+b)^3\sqrt[3]{4ab}}{\sqrt[3]{(a+b)^5}} = 2^{2/3} (a+b)^{4/3} (ab)^{1/3}. $$ Applying AM-GM on $a$ and $b$ we find $a+b \geq 2\sqrt{ab}$. Hence $$ \frac{(a+b)^3}{\sqrt[3]{2(a+b)(a^2+b^2)}} \geq 2^{2/3} (a+b)^{4/3} (ab)^{1/3} \geq 2^{2/3} (2\sqrt{ab})^{4/3} (ab)^{1/3} = 4ab. $$ It follows that the given expression is at least $4(ab+bc+ca) = 24 \geq 12$.

user133281
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A simple and an elegant solution:

first,

$$\sqrt{\frac{a^2+b^2}{2}} \leq \frac{a+b}{2}$$ or $$a^2+b^2 \leq 2 \cdot \frac{(a+b)^2}{4}=\frac{(a+b)^2}{2}$$ So $$2(a+b)(a^2+b^2) \leq 2(a+b)\frac{(a+b)^2}{2}=(a+b)^3$$ now: $$\frac{1}{\sqrt[3]{2(a+b)(a^2+b^2)}} \geq \frac{1}{\sqrt[3]{(a+b)^3}}=\frac{1}{a+b}$$

Our inequality becomes:

$$\frac{(a+b)^3}{(a+b)}+\frac{(b+c)^3}{(b+c)}+\frac{(c+a)^3}{(c+a)} =(a+b)^2+(b+c)^2+(c+a)^2=$$ $$2(a^2+b^2+c^2)+2(ab+bc+ca)=2(a^2+b^2+c^2)+12 \geq 12.$$

Iuli
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We'll prove a stronger inequality: $\sum\limits_{cyc}\frac{(a+b)^3}{\sqrt[3]{2(a+b)(a^2+b^2)}}\geq24$.

Indeed, let $a=\sqrt2x$, $b=\sqrt2y$, $c=\sqrt2z$, $x+y+z=3u$

and $xy+xz+yz=3v^2$, where $v\geq0$.

Hence, $v=1$, $u\geq v$ and we need to prove that $\sum\limits_{cyc}\frac{(x+y)^3}{\sqrt[3]{2(x+y)(x^2+y^2)}}\geq12.$

Now, by AM-GM and Holder we obtain: $$\sum\limits_{cyc}\frac{(x+y)^3}{\sqrt[3]{2(x+y)(x^2+y^2)}}=\sum\limits_{cyc}\frac{3(x+y)^3}{3\sqrt[3]{2(x+y)(x^2+y^2)}}\geq$$ $$\geq\sum\limits_{cyc}\frac{3(x+y)^3}{2+x+y+x^2+y^2}\geq\frac{3\left(\sum\limits_{cyc}(x+y)\right)^3}{\sum\limits_{cyc}(2+x+y+x^2+y^2)}=$$ $$=\frac{4(x+y+z)^3}{3+x+y+z+x^2+y^2+z^2}=\frac{36u^3}{1+u+3u^2-2v^2}=$$ $$=\frac{36u^3}{v^3+uv^2+3u^2v-2v^3}=\frac{36u^3}{3u^2v+uv^2-v^3}\geq12,$$ where the last inequality it's just $(u-v)(3u^2-v^2)\geq0$, which is obvious.

Done!

Michael Rozenberg
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