# Prove that for some real numbers $a$ and $b$: $(a^2+a+1)^{-1/2}+(b^2+b+1)^{-1/2}+4((a+b)^2-2(a+b)+4)^{-1/2} \leq 4$

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)

• $a^2 + a + 1 = 0.75$ when $a = -0.5$ – Larry B. Sep 21 '18 at 19:36
• 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. – LPenguin Sep 23 '18 at 1:01