Given three positive numbers $a,b,c$. Prove that $\prod\limits_{sym}(\!a+\frac{1}{a}\!)\geqq\frac{4}{3}\sum\limits_{sym}\frac{b+c}{a}$ .

Question

Given three positive numbers $a, b, c$. Prove that $$\prod\limits_{sym}\left(\!a+ \frac{1}{a}\!\right)\geqq \frac{4}{3}\sum\limits_{sym}\frac{b+ c}{a}$$

I use discriminant and uvw and find it with the best constant $4\div 3$ .

Titu lemma and Holder inequality both are unsuccessful here, so I need to the help. Thanks a real lot

Answer 1

We need to prove that $$3\prod_{cyc}(a^2+1)\geq4\sum_{cyc}(a^2b+a^2c)$$ or $$3(a^2b^2c^2+1)+3\sum_{cyc}(a^2b^2+a^2)\geq4\sum_{cyc}(a^2b+a^2c).$$ But by AM-GM $$a^2b^2c^2+1\geq2abc.$$ Thus, it's enough to prove that $$6abc+3\sum_{cyc}(a^2b^2+a^2)\geq4\sum_{cyc}(a^2b+a^2c).$$ Now, let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.

Thus, we need to prove that: $$6w^3+3(9v^4-6uw^3+9u^2-6v^2)\geq4(9uv^2-3w^3),$$ which is a linear inequality of $w^3$, which says that it's enough to prove the last inequality for an extreme value of $w^3$, which happens in the following cases.

1. $w^3\rightarrow0^+$.

Let $c\rightarrow0^+$.

Thus, we need to prove that $$3a^2b^2+3a^2+3b^2\geq4(a^2b+ab^2),$$ which is true by AM-GM: $$3a^2b^2+3a^2+3b^2=3\left(\frac{1}{2}a^2b^2+a^2+\frac{1}{2}a^2b^2+b^2\right)\geq$$ $$\geq3\left(2\sqrt{\frac{1}{2}a^2b^2\cdot a^2}+(2\sqrt{\frac{1}{2}a^2b^2\cdot b^2}\right)=3\sqrt2(a^2b+ab^2)>4(a^2b+ab^2);$$ 2. Two variables are equal.

Let $b=a$.

Thus, we need to prove that $$6a^2c+3(a^4+2a^2c^2+2a^2+c^2)\geq8(a^3+a^2c+ac^2)$$ or $$(6a^2-8a+3)c^2-2a^2c+a^2(3a^2-8a+6)\geq0,$$ for which it's enough to prove that $$(6a^2-8a+3)(3a^2-8a+6)\geq a^2$$ or $$(a-1)^4\geq0$$ and we are done!