Algebra (inequalities)

Question

Let $\{a,b\}\subset\mathbb{R}$. Show that: $$a^2+b^2+ \frac{1}{a^2}+ \frac{b}{a} \geq \sqrt{3}$$

I am unable to apply AM-GM inequality in the given problem.

Answer 1

We need to prove that $$b^2+\frac{1}{a}\cdot{b}+a^2+\frac{1}{a^2}-\sqrt3\geq0,$$ for which we need $\Delta\leq0$ or $$\frac{1}{a^2}-4\left(a^2+\frac{1}{a^2}-\sqrt3\right)\leq0$$ or $$4a^2+\frac{3}{a^2}\geq4\sqrt3,$$ which is AM-GM: $$4a^2+\frac{3}{a^2}\geq2\sqrt{4a^2\cdot\frac{3}{a^2}}=4\sqrt3.$$ Done!

Answer 2

it is equivalent to $$\left(ab+\frac{1}{2}\right)^2+\left(a^2-\frac{\sqrt{3}}{2}\right)^2\geq 0$$

Answer 3

One could do as follows as well, if some calculus were allowed.

Consider the function $ f(a,b) = a^2 + b^2 + \frac{1}{a^2} + \frac{b}{a}$. As we are looking for a lower bound, we could calculate $ \frac{\partial f}{\partial b} = 2b + \frac{1}{a}$ and, upon setting to zero, look along the domain restriction $$ b = -\frac{1}{2a}$$ only. Substituting yields the expression $$a^2 + \frac{3}{4a^2} = \frac{2a^2 + \frac{3}{2a^2}}{2} $$ after which we can apply the AM-GM inequality $$\frac {x+y}{2} \geq \sqrt {x y}$$ upon setting $x=2a^2$ and $y = \frac{3}{2a^2} $, concluding that the given expression is equal or greater than $$ \sqrt{ 2a^2 \frac{3}{2a^2}} = \sqrt{3}$$