<s>[INST] 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 
 [/INST] your inequality is equivalent to $${\frac { \left( {a}^{2}-ab+{b}^{2} \right) ^{2}}{{b}^{2}{a}^{2}}}\geq 0$$ and this is true for all $$a,b>0$$
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