Show $\forall a,b\in\mathbb{R}^*\quad 3+\frac{a^2}{b^2}+\frac{b^2}{a^2}\ge 2 (\frac{a}{b}+\frac{b}{a})$

Question

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

@Educ Just asking. Maybe I'm wrong. Could it more general if we just prove $$3+x^2+y^2\geq 2(x+y),$$ where $x,y\in\mathbb{R}^*$? — Juniven Acapulco, Dec 29 '16 at 10:24

score 2 · Answer 1 · answered Dec 29 '16 at 10:21

2

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.

answered Dec 29 '16 at 10:21

DeepSea

77,651

score 1 · Answer 2 · answered Dec 29 '16 at 10:20

1

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 $$a,b>0$$

answered Dec 29 '16 at 10:20

Dr. Sonnhard Graubner

95,283

2

its true for all $a,b\ne0$ – JMP Dec 29 '16 at 10:38
but the original problem demands $$a,b>0$$ – Dr. Sonnhard Graubner Dec 29 '16 at 10:40
well, you've solved it for a larger range then – JMP Dec 29 '16 at 10:44
The original problem asks for $a$ and $b$ in $\mathbb R^*$, which often stands for $\mathbb R\setminus{0}$, thus $a<0$ or $b<0$ should also be considered. – Did Dec 29 '16 at 11:15
Agreeable solution. It is valid for all $ab\ne 0$ – Piquito Dec 29 '16 at 11:43
This is hardly different from the solution posted by OP. Just plug $t=a/b+b/a$ in $(t-1)^2\ge0$. – Cave Johnson Dec 30 '16 at 11:53
yes this is another solution by me, is this forbidden? – Dr. Sonnhard Graubner Dec 30 '16 at 11:54
I mean this is not another solution. It's exactly the same solution posted by OP. – Cave Johnson Dec 31 '16 at 04:07

Show $\forall a,b\in\mathbb{R}^*\quad 3+\frac{a^2}{b^2}+\frac{b^2}{a^2}\ge 2 (\frac{a}{b}+\frac{b}{a})$

2 Answers2