Inequality

[man111]

If $a,b,c\in\mathbb{R^{+}}\;,a+b+c=1$ Then prove that

$\displaystyle\left(a+\frac{1}{b}\right).\left(b+\frac{1}{c}\right).\left(c+\frac{1}{a}\right)\geq \left(\frac{10}{3}\right)^3$

[arqady]

By AM-GM we obtain: $abc\leq\frac{1}{27}$.

$\displaystyle\left(a+\frac{1}{b}\right)\left(b+\frac{1}{c}\right)\left(c+\frac{1}{a}\right)\geq \left(\frac{10}{3}\right)^3\Leftrightarrow\prod_{cyc}(ab+1)\geq\left(\frac{10}{3}\right)^3abc$.
But by Holder $\prod\limits_{cyc}(ab+1)\geq\left(\sqrt[3]{a^2b^2c^2}+1\right)^3$.
Id est, it remains to prove that $\sqrt[3]{a^2b^2c^2}+1\geq\frac{10}{3}\sqrt[3]{abc}$, which is $\left(3\sqrt[3]{abc}-1\right)\left(\sqrt[3]{abc}-3\right)\geq0$,
which is true because $abc\leq\frac{1}{27}$.

Следующее более строгое неравенство тоже верно.
Пусть $a$, $b$ и $c$ положительны и $a+b+c=1$. Докажите, что

$$\displaystyle\left(a+\frac{1}{b}\right)\left(b+\frac{1}{c}\right)\left(c+\frac{1}{a}\right)\geq \frac{1000}{9}(a^2+b^2+c^2)$$

[man111]

thanks arqady