$\sum \frac{a^2+b^2}{c^2}-\frac{7}{2} \geq \sqrt[3]{\prod \left(\frac{a}{b+c}+\frac{1}{3}\right)}$ for a,b,c real

Question

Let: $a,b,c>0$. Prove that: $$\frac{a^2+b^2}{c^2}+\frac{b^2+c^2}{a^2}+\frac{c^2+a^2}{b^2}-\frac{7}{2}\geq \sqrt[3]{\left(\frac{a}{b+c}+\frac{1}{3}\right)\left(\frac{b}{c+a}+\frac{1}{3}\right)\left(\frac{c}{a+b}+\frac{1}{3}\right)}.$$

Here are my solution:

Can anyone show me a better proof not using $ln$ or some hard method, just use AM-GM, pqr or classic inequality.

And I still think what if $a,b,c$ are real numbers, is the inequality still true?

Answer 1

For real numbers it's wrong. Try $(a,b,c)=(6,3,-1).$

$uvw$ kills your inequality and we can create something stronger.

We'll prove a stronger inequality.

Let $a$, $b$ and $c$ be positive numbers. Prove that: $$\frac{a^2+b^2}{c^2}+\frac{a^2+c^2}{b^2}+\frac{b^2+c^2}{a^2}+48\geq\frac{36(a+b+c)}{\sqrt[3]{(a+b)(a+c)(b+c)}}.$$

It's stronger because by AM-GM $$\frac{36(a+b+c)}{\sqrt[3]{(a+b)(a+c)(b+c)}}-48\geq\frac{7}{2}+3\sqrt[3]{\prod_{cyc}\left(\frac{a}{b+c}+\frac{1}{3}\right)}.$$ Indeed, let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.

Thus, we need to prove that: $$\frac{(a^2+b^2+c^2)(a^2b^2+a^2c^2+b^2c^2)-3a^2b^2c^2}{a^2b^2c^2}+48\geq\frac{36(a+b+c)}{\sqrt[3]{(a+b+c)(ab+ac+bc)-3abc}}$$ or $f(u)\geq0,$ where $$f(u)=(3u^2-2v^2)(9v^4-6uw^3)-w^6+16w^6-\frac{36uw^6}{\sqrt[3]{9uv^2-w^3}}.$$ But by Schur $$f'(u)=6u(9v^4-6uw^3)-6w^3(3u^2-2v^2)-36w^6\left(\frac{1}{\sqrt[3]{9uv^2-w^3}}-\frac{3uv^2}{\sqrt[3]{(9uv^2-w^3)^4}}\right)=$$ $$=54uv^4-54u^2w^3+12v^2w^3-\frac{36w^6(6uv^2-w^3)}{\sqrt[3]{(9uv^2-w^3})^4}\geq$$ $$\geq12w^3\left(v^2-\frac{3w^3(6uv^2-w^3)}{\sqrt[3]{8v^{12}}}\right)=\frac{12w^3(16v^6-18uv^2w^3+3w^6)}{16v^4}\geq$$ $$\geq\tfrac{3w^3(8v^6-9uv^2w^3+w^6)}{2v^4}\geq\tfrac{3w^3(8v^6-9uv^2w^3+4uv^2w^3-3v^6)}{2v^4}=\tfrac{15v^2w^3(v^4-uw^3)}{2v^4}\geq0.$$

Thus, $f$ increases, which by $uvw$ says that it's enough to prove $f(u)\geq0$ for equality case of two variables.

Let $b=c=1$.

Thus, we need to prove that: $$\frac{2}{a^2}+2+2a^2+48\geq\frac{36(a+2)}{\sqrt[3]{2(a+1)^2}},$$ which is smooth.

The following stronger inequality is also true.

Let $a$, $b$ and $c$ be positive numbers. Prove that: $$\frac{a^2+b^2}{c^2}+\frac{a^2+c^2}{b^2}+\frac{b^2+c^2}{a^2}+246\geq\frac{168(a+b+c)}{\sqrt[3]{(a+b)(a+c)(b+c)}},$$

but my proof of this statement is very ugly.