Prove an inequality concerning $\sqrt[3]{4a^3+4b^3}+\sqrt[3]{4b^3+4c^3}+\sqrt[3]{4c^3+4a^3}$

Question

Let $a,b,c$ be positive. I need to prove

$\sqrt[3]{4a^3+4b^3}+\sqrt[3]{4b^3+4c^3}+\sqrt[3]{4c^3+4a^3}\leq \dfrac{4a^2}{a+b}+\dfrac{4b^2}{b+c}+\dfrac{4c^2}{c+a}$

Thanks!

Answer 1

Observe that

$$\sum \frac{4a^2 - 4b^2} { a+b} = \sum (4a-4b) = 0. $$

Hence, we have

$$\sum \frac{4a^2} { a+b} = \sum \frac{4b^2} {a+b}$$

Hence, we may rewrite the RHS and make it more symmetric, namely

$$\sum \sqrt[3]{4a^3+4b^3} \leq \sum \frac{4a^2} {a+b} = \sum \frac{ 2a^2+2b^2} { a+b}.$$

Now, we can break it up into the individual terms, and just show that

$$ \frac{ a^3+b^3} {2} \left( \frac{a+b}{2} \right)^3 \leq \left( \frac{a^2+b^2} {2} \right)^3$$

This is true because

[Newton] $$S_1 S_3 \leq S_2^2$$
[Maclaurin] $$S_1^2 \leq S_2$$

Answer 2

I know that Calvin already answered your question, but I was bored and resoluted to prove it only with elementary tools. So here goes my proof for what is worth:

Manipulate the RHS to find symmetry $$\frac{4a^2}{a+b}+\frac{4b^2}{b+c}+\frac{4c^2}{c+a}$$ $$4\left(\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+a}\right)$$ $$4\left(a+b+c-\frac{ab}{a+b}-\frac{bc}{b+c}-\frac{ac}{c+a}\right)$$ $$2\left(a+b-\frac{2ab}{a+b}+b+c-\frac{2bc}{b+c}+a+c-\frac{2ac}{c+a}\right)$$ $$2\left(\frac{a^2+b^2}{a+b}+\frac{b^2+c^2}{b+c}+\frac{a^2+c^2}{c+a}\right)$$ Observe that $$0\le(2x-y)^2$$ $$0\le4x^2-4xy+y^2$$ $$3x(y-x)\le x^2-xy+y^2$$ $$3x(y^2-x^2)\le x^3+y^3$$ Substituting

$x=ab,\,y=a^2+b^2$ $$3ab(a^2b^2+a^4+b^4)\le(a^2+b^2)^3+a^3b^3$$ $$(a^2+b^2)^3+8a^3b^3+3ab(a^4+b^4-2a^2b^2))\le2(a^2+b^2)^3$$ $$(a^2+b^2+2ab)((a^2+b^2)^2-2ab(a^2+b^2)+4a^2b^2)+3ab(a^2-b^2)^2\le2(a^2+b^2)^3$$ $$(a+b)^2((a^2+b^2)^2-2ab(a^2+b^2)+4a^2b^2)+3ab(a+b)^2(a-b)^2\le2(a^2+b^2)^3$$ $$(a+b)^2((a^2+b^2)^2-2ab(a^2+b^2)+4a^2b^2+3ab(a-b)^2)\le2(a^2+b^2)^3$$ $$(a+b)^2((a^2+b^2)^2+ab(a^2+b^2)-2a^2b^2)\le2(a^2+b^2)^3$$ $$(a+b)^2((a^4+b^4+ab(a^2+b^2))\le2(a^2+b^2)^3$$ $$(a+b)^2(a+b)(a^3+b^3)\le2(a^2+b^2)^3$$ $$4(a^3+b^3)(a+b)^3\le8(a^2+b^2)^3$$ $$\sqrt[3]{4(a^3+b^3)}\le\frac{2(a^2+b^2)}{a+b}$$ $$\sqrt[3]{4(b^3+c^3)}\le\frac{2(b^2+c^2)}{b+c}$$ $$\sqrt[3]{4(a^3+c^3)}\le\frac{2(a^2+c^2)}{a+c}$$ $$\sqrt[3]{4(a^3+b^3)}+\sqrt[3]{4(b^3+c^3)}+\sqrt[3]{4(a^3+c^3)}\le2\left(\frac{a^2+b^2}{a+b}+\frac{b^2+c^2}{b+c}+\frac{a^2+c^2}{c+a}\right)$$ $$\sqrt[3]{4(a^3+b^3)}+\sqrt[3]{4(b^3+c^3)}+\sqrt[3]{4(a^3+c^3)}\le\frac{4a^2}{a+b}+\frac{4b^2}{b+c}+\frac{4c^2}{c+a}$$ I like how it looks like a cup.