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Let $a,b,c,d$ be positive real numbers such that $a+b+c+d=1$, then prove that $$S=\sqrt{4a+1}+\sqrt{4b+1}+\sqrt{4c+1}+\sqrt{4d+1} < 6$$

Source : Inequalities (Page Number 197)

I used Cauchy-Schwarz Inequality on these two sets : $\{\sqrt{4a+1},\sqrt{4b+1},\sqrt{4c+1},\sqrt{4d+1}\}$ and $\{1,1,1,1\}$ to get:

$$S^2 \leq \big(4a+1+4b+1+4c+1+4d+1\big)(4) = \big(4(a+b+c+d) + 4\big)(4)=32$$

$$\implies S \leq \sqrt{32} = 4\sqrt{2} \approx 5.6568$$

So, I am able to prove a somewhat "sharper" inequality than the one given.

Just out of curiosity, I wanted to know if there is a different and preferably more elegant method of tackling this problem.

How to actually prove the original inequality ?

Thanks in advance ! :)

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Also we have the following $$\sqrt{4a+1}<2a+1$$ Hence, $$\sum_{cyc}\sqrt{4a+1}<\sum_{cyc}(2a+1)=6$$

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  • $\begingroup$ Thanks a lot!!!! That's really ingenious!!! Simplest approach possible... $\endgroup$ – user399078 Feb 10 '17 at 14:29
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Let $f(x)=\sqrt{x}$. Hence, $f$ is a concave function.

Thus, by Jensen $$\sum_{cyc}\sqrt{4a+1}\leq\sqrt{4\sum_{cyc}(4a+1)}=\sqrt{32}<6$$

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  • $\begingroup$ Thank you for a different approach, but I have already proved that $S\leq\sqrt{32}$.... Is there some way to exactly prove $S<6$ without proving a sharper inequality...??? $\endgroup$ – user399078 Feb 10 '17 at 14:26

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