Show that problem and that equality if fulfilled iff a=b

Question

If $a, b >0$, show that $$ \frac{a}{b^2} + \frac{b}{a^2} \geq \frac{1}{a} +\frac{1}{b},$$ and that equality is fulfilled if and only if $a = b.$

I tried using elementary consequences of order axioms, but I had trouble applying them. If you can explain this to me, it would be great.

Thanks.

Answer 1

Hint:   multiply by $\,a^2b^2 \gt 0\,$, then $\;a^3+b^3 \ge ab^2+a^2b \iff (a-b)(a^2-b^2) \ge 0\,$.

Answer 2

First,

$$\frac{a}{b^2} + \frac{b}{a^2}=\frac{a^3+b^3}{a^2b^2}\ge \frac{a+b}{ab}$$

since $a,b>0$ we can deduce

$$\frac{a^3+b^3}{ab}\ge a+b$$

And since $(a+b)(a^2-ab+b^2)=a^3+b^3$ we get

$$a^2-ab+b^2\ge ab\Leftrightarrow (a-b)^2\ge 0$$

And equality iff $a=b$

Answer 3

$$\frac13\frac{a}{b^2}+\frac23\frac{b}{a^2}\ge \sqrt[3]{\frac{1}{a^3}}=\frac1a$$ by AM-GM. Adding up symmetrically and we are done.

Answer 4

Using Cauchy-Schwarz inequality, we have: $\left(\dfrac{1}{a}+\dfrac{1}{b}\right)^2= \left(\dfrac{\sqrt{b}}{a}\cdot \dfrac{1}{\sqrt{b}}+\dfrac{\sqrt{a}}{b}\cdot \dfrac{1}{\sqrt{a}}\right)^2\le \left(\dfrac{b}{a^2}+\dfrac{a}{b^2}\right)\left(\dfrac{1}{b}+\dfrac{1}{a}\right)\implies \dfrac{1}{a}+\dfrac{1}{b}\le \dfrac{a}{b^2}+\dfrac{b}{a^2}$

Answer 5

$(a,b)$ and $\left(\frac{1}{a^2},\frac{1}{b^2}\right)$ they are opposite ordered.

Thus, by Rearrangement we obtain: $$\frac{a}{b^2}+\frac{b}{a^2}=a\cdot\frac{1}{b^2}+b\cdot\frac{1}{a^2}\geq a\cdot\frac{1}{a^2}+b\cdot\frac{1}{b^2}=\frac{1}{a}+\frac{1}{b}.$$ The equality occurs iff $\left(\frac{1}{a^2},\frac{1}{b^2}\right)=\left(\frac{1}{b^2},\frac{1}{a^2}\right)$, id est, for $a=b$.

Also, by C-S we obtain: $$\frac{a}{b^2}+\frac{b}{a^2}=\frac{a^2}{ab^2}+\frac{b^2}{ba^2}\geq\frac{(a+b)^2}{ab^2+ba^2}=\frac{1}{a}+\frac{1}{b}.$$