2
$\begingroup$

Proof the inequality for some real numbers $a$, $b$: $$\frac{1}{\sqrt{a^2+a+1}}+\frac{1}{\sqrt{b^2+b+1}}+\frac{4}{\sqrt{(a+b)^2-2(a+b)+4}} \leq 4$$


My idea was: $(0+0+1)(a^2+a+1)\geq 1$ (CBS) and $(0+0+1)(b^2+b+1)\geq 1$ (CBS)


But I don't know if that is correct. Please help. :)

$\endgroup$
  • $\begingroup$ $a^2 + a + 1 = 0.75$ when $a = -0.5$ $\endgroup$ – Larry B. Sep 21 '18 at 19:36
  • $\begingroup$ Something along these lines might work. $a^2+a+1$, $b^2+b+1$, $(a+b)^2-2(a+b)+4=(a+b-1)^2+3$ are all increasing functions with respect to $a$ and $b$. The infimum of the set of values the LHS can take is therefore $\lim_{a\rightarrow \infty }\lim_{b\rightarrow \infty }\left ( \frac{1}{\sqrt{a^{2}+a+1}}+\frac{1}{\sqrt{b^{2}+b+1}} + \frac{4}{\sqrt{\left ( a+b \right )^{2}-2\left ( a+b \right )+4}} \right ) = 0$. Less than $4$, so simply choose $a,b$ large enough. This strengthens the result to the RHS being any non-zero positive number. $\endgroup$ – LPenguin Sep 23 '18 at 1:01

Your Answer

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

Browse other questions tagged or ask your own question.