Show $\forall a,b\in\mathbb{R}^*\quad 3+\frac{a^2}{b^2}+\frac{b^2}{a^2}\ge 2 (\frac{a}{b}+\frac{b}{a})$ Let $a,b \in \mathbb{R}^*$ Show that :
$$3+\frac{a^2}{b^2}+\frac{b^2}{a^2}\ge 2\left(\frac{a}{b}+\frac{b}{a}\right)$$
Let $a,b\in\mathbb{R}^*$
let $t=\dfrac{a}{b}+\dfrac{b}{a}$ , we've :
$$
\begin{aligned}
3+\frac{a^2}{b^2}+\frac{b^2}{a^2}\ge 2\left(\frac{a}{b}+\frac{b}{a}\right) &\iff
3+\frac{a^2}{b^2}+\frac{b^2}{a^2}-2\left(\frac{a}{b}+\frac{b}{a}\right)\\
&\iff t^{2}-2t+1\geq 0\\
&\iff \left(t-1\right)^{2}\geq 0
\end{aligned}$$
since $\forall t\in \mathbb{R}\quad (t-1)^{2}\geq 0$ holds then also  $$\forall a,b\in \mathbb{R}^*\quad 3+\frac{a^2}{b^2}+\frac{b^2}{a^2}\ge 2\left(\frac{a}{b}+\frac{b}{a}\right) $$ holds
Am i right ? is there any other ways 

We have: $1+\dfrac{a^2}{b^2} \ge \dfrac{2a}{b}, 1+\dfrac{b^2}{a^2} \ge \dfrac{2b}{a}$. Adding these we have the desire inequality. Thus the inequality is strict.
