Let $a,b,c > 0$ such that $abc \geq 1$. Prove that: $$ \frac{a^2}{b} + \frac{b^2}{c} + \frac{c^2}{a} \geq \frac{1}{a} + \frac{1}{b} + \frac{1}{c}$$
I have tried Cauchy-Schwarz inequality: $$LHS \geq a+b+c$$
And also: $$RHS \leq ab+bc+ca$$
So I need to prove: $$a+b+c \geq ab+bc+ca \Leftrightarrow (a+b+c)^2\geq(a+b+c)(ab+bc+ca)$$ (1)
But also I got: $$a+b+c \geq 3 \sqrt[3]{abc} \geq 3$$
So I want to change (1) into: $$(a+b+c)^2 \geq 3(ab+bc+ca)$$
I also think Schur and pqr method but they are to complicated
Best Answer
Applying the Cauchy–Schwarz inequality, you have :
$$ \begin{align}\small {\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}≥\frac {\left(\dfrac ab+\dfrac bc+\dfrac ca\right)^2}{\dfrac 1b+\dfrac 1c+\dfrac 1a}}\end{align} $$
Thus, it is enough to show that :
$$ \begin{align}\frac ab+\frac bc+\frac ca≥\frac 1a+\frac 1b+\frac 1c\end{align} $$
Finally, using the AM-GM inequality, we have :
$$ \begin{align}\begin{cases}\dfrac ab+\dfrac ab+\dfrac ca≥\dfrac {3\sqrt [3]{abc}}{b}≥\dfrac 3b\\ \dfrac ca+\dfrac ca+\dfrac bc≥\dfrac {3\sqrt [3]{abc}}{a}≥\dfrac 3a\\ \dfrac bc+\dfrac bc+\dfrac ab≥\dfrac {3\sqrt [3]{abc}}{c}≥\dfrac 3c\end{cases}\end{align} $$
By summing up the inequalities side-by-side, we conclude that :
$$ \begin{align}\frac ab+\frac bc+\frac ca≥\frac 1a+\frac 1b+\frac 1c\end{align} $$
which completes the proof .