2
$\begingroup$

$a,b$ 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 ?

$\endgroup$
2
$\begingroup$

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 $5x^2+4 \ge 4\sqrt5 |x|> 6|x|$.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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

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

$\endgroup$
  • $\begingroup$ Nice, But the fact is, This question was given to to my friend To be solved in class. So, maybe, they might expect a easy solution. But, Thanks for your approach too. I will try to gain this result through my efforts $\endgroup$ – Dinesh Oct 13 '14 at 0:30
  • 1
    $\begingroup$ Is this inequality supposed to be obvious? $\endgroup$ – Cheerful Parsnip Oct 13 '14 at 1:20
0
$\begingroup$

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!

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.