An Inequality; $a^2+b^2=1$

Question

$a,b$ are tho real numbers such that $a^2+b^2=1$. To prove that ;

$$\dfrac{1}{a^2+1}+\dfrac{1}{b^2+1}+\dfrac{1}{ab+1}\geq\dfrac{3}{1+\cfrac{(a+b)^2}{4}}$$

When I first saw this question, I thought of applying Titu's Lemma, to get $$\dfrac{1}{a^2+1}+\dfrac{1}{b^2+1}+\dfrac{1}{ab+1}\geq\dfrac{9}{1+1+1+a^2+b^2+ab}=\dfrac{9}{3+(a+b)^2-ab}OR=\dfrac{9}{4+ab}$$

Now, from hereI am confused, how to proceed. Can anybody kindly help me over this problem ?

Answer 1

Let $a = \cos(t)$, $b = \sin(t)$. Then (using Maple) the inequality simplifies to

$$ {\frac { \left( 1 + \dfrac32\,\sin \left( 2\,t \right) +\dfrac54\, \sin^2 \left( 2\,t \right) \right) \left( \cos \left( t \right) -\sin \left( t \right) \right) ^{2}}{ \left( \cos^2 \left( t \right) +1 \right) \left( \sin^2 \left( t \right) +1 \right) \left( \cos \left( t \right) \sin \left( t \right) +1 \right) \left((\cos(t)+\sin(t))^2+4\right) }} \ge 0 $$

Answer 2

Your partial attempt above using Titu's Lemma cannot succeed because if I put $a = \sqrt{3}/2$ and $b = 1/2$, then $9/(4 + ab) < 3 / (1 + (a+b)^2/4)$.

I would also go with the trigonometric solution above though some care is needed to show that the expression above is positive when one of $a$ or $b$ is negative.

Answer 3

This is more of a step-by-step version of the trigonometric soln already posted. With $a=\cos t, \; b = \sin t$, the inequality is $$\frac1{\sin^2t+1}+\frac1{\cos^2t+1}+\frac1{\frac12 \sin 2t+1} \ge \frac{12}{5+\sin 2t}$$

$$LHS = \frac{3}{\frac14\sin^22t + 2}+\frac1{\frac12\sin 2t+1}$$ so with $x = \sin 2t\in [-1, 1]$, the inequality can be written as $$\frac{12}{8+x^2}+\frac2{2+x}\ge \frac{12}{5+x} \iff (1-x)(4+6x+5x^2) \ge 0$$

which is obvious as $5x^2+4 \ge 4\sqrt5 |x|> 6|x|$.

Answer 4

We need to prove that $$\frac{1}{2a^2+b^2}+\frac{1}{a^2+2b^2}+\frac{1}{a^2+ab+b^2}\geq\frac{12}{5a^2+2ab+5b^2}$$ or $$(a-b)^2(a^4+3a^3b+7a^2b^2+3ab^3+b^4)\geq0,$$ which is obvious.

Done!