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Let $a,b,c$ be the nonnegative real numbers such that $a+b+c=1$. Prove that $$\sqrt{a+\frac{(b-c)^2}4}+\sqrt b+\sqrt c\le\sqrt3$$

I first wrote $a$ as $1-b-c$ and substituted it in main inequality $$\sqrt{4(1-b-c)+(b-c)^2}+2\sqrt b+2\sqrt c\le2\sqrt 3$$ I tried to find some relation between this inequality and QAGH inequalities, but I couldn't. Then I come up with an idea to square main inequality several times to cancel square roots and then to reduce it to sum of squares, but I am sure there is a better way to prove it. What is the best way to prove it?

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I use Cauchy-Schwarz inequality we have $$\left(\sqrt{a+\dfrac{(b-c)^2}{4}}+\sqrt{b}+\sqrt{c}\right)^2 \le \left(a+\dfrac{(b-c)^2}{4}+\dfrac{(\sqrt{b}+\sqrt{c})^2}{2}\right)(1+2) $$ $$\Longleftrightarrow \left(a+\dfrac{(b-c)^2}{4}+\dfrac{(\sqrt{b}+\sqrt{c})^2}{2}\right)\le 1$$ since $1-a=b+c$ $$\Longleftrightarrow \dfrac{(b-c)^2}{4}+\dfrac{(\sqrt{b}+\sqrt{c})^2}{2}\le b+c$$ $$\Longleftrightarrow \dfrac{(b-c)^2}{4}\le\dfrac{(\sqrt{b}-\sqrt{c})^2}{2}$$ $$\Longleftrightarrow (\sqrt{b}+\sqrt{c})^2\le 2$$ It is clear,Because use Cauchy-schwarz inequality $$b+c\le 1\Longrightarrow (\sqrt{b}+\sqrt{c})^2\le[1+1](b+c)$$

math110
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  • As far I know, we can use Cauchy-Schwarz inequality iff numbers are linearly dependent. I cannot understand how you applied this inequality here. –  Mar 28 '15 at 10:05
  • do you know $(X^2+Y^2)(Z^2+W^2)\ge (XZ+YW)^2?$ – math110 Mar 28 '15 at 10:17
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    The Cauchy-Schwarz inequality is true regardless of whether the numbers are linearly dependent or not. I think what you mean is that equality holds iff the numbers you're using Cauchy on are linearly dependent, but you don't actually need that here. – Some Math Student Mar 28 '15 at 10:18
  • if you want to rely on mean inequalities, use $X+Y=m(X/m)+n(Y/n)\le (m+n)\sqrt{\frac{m(X/m)^2+n(Y/n)^2}{m+n}}=\sqrt{m+n}\sqrt{X^2/m+Y^2/n}$. – Lutz Lehmann Mar 28 '15 at 11:59