For $a,b,c>0$ . Prove $ (a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})\geq 9+8\frac{(a-b)^{2}+(b-c)^{2}+(c-a)^{2}}{(a+b+c)^{2}}$ For $a,b,c>0$ . Prove
$$(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\geq 9+8\frac{(a-b)^{2}+(b-c)^{2}+(c-a)^{2}}{(a+b+c)^{2}}$$
 A: Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Hence, we need to prove that
$$\frac{(a+b+c)(ab+ac+bc)}{abc}\geq9+\frac{16(a^2+b^2+c^2-ab-ac-bc)}{(a+b+c)^2}$$ or
$$\frac{9uv^2}{w^3}\geq9+\frac{16(u^2-v^2)}{u^2},$$
which is $f(w^3)\geq0$, where $f$ is an decreasing function.
Thus, it's enough to prove our inequality for a maximal value of $w^3$, which happens for equality case of two variables.
Let $b=c=1$.
Hence, we need to prove that $(a-1)^2(a-2)^2\geq0.$
Done!
Another way.
$$(a-b)^2(a-c)^2(b-c)^2=27(3u^2v^4-4v^6-4u^3w^3+6uv^2w^3-w^6)\geq0,$$
which gives
$$3uv^2-2u^3-2\sqrt{(u^2-v^2)^3}\leq w^3\leq3uv^2-2u^3+2\sqrt{(u^2-v^2)^3}.$$
we need to prove that
$$\frac{9uv^2}{w^3}\geq9+\frac{16(u^2-v^2)}{u^2}$$ or
$$\frac{9uv^2}{w^3}\geq\frac{25u^2-16v^2}{u^2}$$ or
$$w^3\leq\frac{9u^3v^2}{25u^2-16v^2},$$ for which it's enough to prove that
$$3uv^2-2u^3+2\sqrt{(u^2-v^2)^3}\leq\frac{9u^3v^2}{25u^2-16v^2}$$ or
$$25u^3-49u^3v^2+24uv^4\geq(25u^2-16v^2)\sqrt{(u^2-v^2)^3}$$ or
$$u(25u^2-24v^2)\geq(25u^2-16v^2)\sqrt{(u^2-v^2)}$$ or
$$u^2(25u^2-24v^2)^2\geq(25u^2-16v^2)^2(u^2-v^2)$$ or
$$(15u^2-16v^2)^2\geq0.$$
Done again!
Also we have the following solution.
$$(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-9-8\cdot\frac{(a-b)^{2}+(b-c)^{2}+(c-a)^{2}}{(a+b+c)^{2}}=$$
$$=\frac{\sum\limits_{cyc}c(a-b)^2(a+b-3c)^2}{abc(a+b+c)^2}\geq0.$$
