If $a,b,c>0$ and $abc≥1$, prove that $ \frac{a^2}{b} + \frac{b^2}{c} + \frac{c^2}{a} \geq \frac{1}{a} + \frac{1}{b} + \frac{1}{c}$

Question

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

Answer 1

Because by AM-GM we obtain: $$\sum_{cyc}\frac{a^2}{b}=\frac{1}{21}\sum_{cyc}\left(\frac{8a^2}{b}+\frac{11b^2}{c}+\frac{2c^2}{a}\right)\geq\sum_{cyc}\sqrt[21]{\left(\frac{a^2}{b}\right)^8\left(\frac{b^2}{c}\right)^{11}\left(\frac{c^2}{a}\right)^2}=$$ $$=\sum_{cyc}\sqrt[21]{\frac{a^{14}b^{14}}{c^7}}=\sum_{cyc}\sqrt[3]{\frac{a^2b^2}{c}}=\sum_{cyc}\sqrt[3]{\frac{a^2b^2c^2}{c^3}}\geq\sum_{cyc}\frac{1}{c}.$$ We can get coefficients $8$, $11$ and $2$ by the Weighted AM–GM:

Let $a_i>0$, $\alpha_i>0$ such that $\alpha_1+\alpha_2+...+\alpha_n=1$. Prove that: $$\alpha_1a_1+\alpha_2a_2+...+\alpha_na_n\geq a_1^{\alpha_1}a_2^{\alpha_2}...a_n^{\alpha_n}.$$

This statement follows immediately by Jensen from the concavity of $\ln$.

Now, since $1\leq abc$, we need to prove that $$\sum_{cyc}\frac{a^2}{b}\geq\sqrt[3]{a^2b^2c^2}\sum_{cyc}\frac{1}{c}.$$ We'll try to find positive values of $\alpha$, $\beta$ and $\gamma$ such that $\alpha+\beta+\gamma=1$ and the inequality $$\alpha\frac{a^2}{b}+\beta\frac{b^2}{c}+\gamma\frac{c^2}{a}\geq\frac{\sqrt[3]{a^2b^2c^2}}{c}$$ would be true by AM-GM, for which we need $$\left(\frac{a^2}{b}\right)^{\alpha}\left(\frac{b^2}{c}\right)^{\beta}\left(\frac{c^2}{a}\right)^{\gamma}=\frac{\sqrt[3]{a^2b^2c^2}}{c}$$ and we obtain the following system: $$2\alpha-\gamma=\frac{2}{3},$$ $$2\beta-\alpha=\frac{2}{3}$$ and $$\alpha+\beta+\gamma=1,$$ which gives $$(\alpha,\beta,\gamma)=\left(\frac{8}{21},\frac{11}{21},\frac{2}{21}\right).$$

Answer 2

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 :

$$ \begin{align}\frac ab+\frac bc+\frac ca≥\frac 1a+\frac 1b+\frac 1c\end{align} $$

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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 .