For all $a, b, c>0$ and $a+b+c=3$ Prove that $$ \sqrt{\frac{a}{b+3} } +\sqrt{\frac{b}{c+3} } +\sqrt{\frac{c}{a+3} } \leq \frac{3}{2} $$ I tried cauchy-schwarz inequality for the L. H. S like and I get $ [\left( \sqrt{a} \right) ^{2}+\left( \sqrt{b} \right) ^{2}+\left( \sqrt{c} \right) ^{2}]\left[ \left( \frac{1}{\sqrt{b+3} } \right) ^{2}+\left( \frac{1}{\sqrt{c+3} } \right) ^{2}+\left( \frac{1}{\sqrt{a+3} } \right) ^{2}\right] \geq \left( \frac{\sqrt{a} }{\sqrt{b+3} } +\frac{\sqrt{b} }{\sqrt{c+3} } +\frac{\sqrt{c} }{\sqrt{a+3} } \right) ^{2}$ Then I get by AM-GM the maximum value of $abc=1$ and the inequality have the value 3... How I can prove inequality?.

  • $\begingroup$ @user236182 yes. I edited the question $\endgroup$
    – user260699
    Nov 27 '15 at 18:37
  • $\begingroup$ The C-S approach won't work. The $LHS$ that you got is too large. Let $(a,b,c)=(2,0.5,0.5)$. Then $\sqrt{[\left( \sqrt{a} \right) ^{2}+\left( \sqrt{b} \right) ^{2}+\left( \sqrt{c} \right) ^{2}]\left[ \left( \frac{1}{\sqrt{b+3} } \right) ^{2}+\left( \frac{1}{\sqrt{c+3} } \right) ^{2}+\left( \frac{1}{\sqrt{a+3} } \right) ^{2}\right]} >\frac{3}{2}$ $\endgroup$
    – user236182
    Nov 27 '15 at 18:47
  • $\begingroup$ @user236182 so could you tell me a little hint? $\endgroup$
    – user260699
    Nov 27 '15 at 19:07
  • $\begingroup$ @user260699 Could you share the source of this problem? Thanks. $\endgroup$
    – Kay K.
    Nov 29 '15 at 15:01

First, we use Cauchy-Schwartz inequality: $$ A^2=\left(\sqrt{\frac{a}{b+3} } +\sqrt{\frac{b}{c+3} } +\sqrt{\frac{c}{a+3} }\right)^2 =\\ \left(\sqrt{\frac{a}{(a+3)(b+3)}(a+3) } +\sqrt{\frac{b}{(b+3)(c+3)} (b+3)} +\sqrt{\frac{c}{(c+3)(a+3)}(c+3) }\right)^2 \leq \\ \left({\frac{a}{(a+3)(b+3)} } +{\frac{b}{(b+3)(c+3)} } +{\frac{c}{(c+3)(a+3)} }\right)\times (a+3+b+3+c+3) . $$ Since $a+3+b+3+c+3=12$, it is enough to prove: $$ \left({\frac{a}{(a+3)(b+3)} } +{\frac{b}{(b+3)(c+3)} } +{\frac{c}{(c+3)(a+3)} }\right)\leq \frac 3{16}. (\star) $$ From this step on, I went the ugly way, since the manipulation was not so frightening. With a simple and nice proof of this step, the whole proof would become much easier.

Anyway, after the simple multiplications, we get to the following inequality: $$ 18-7(ab+ac+bc)+3abc\geq 0. $$ This can be further simplified, using the fact such as $a^2+b^2+c^2=9-2(ab+ac+bc)$ and using the famous following identity: $$ a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-ac-bc). $$ We can further simplify the inequality to the following: $$ a^3+b^3+c^3\geq a^2+b^2+c^2. $$ This follows from the following steps:

  1. Cauchy-Schwartz: $(a+b+c)(a^3+b^3+c^3)\geq (a^2+b^2+c^2)^2 \implies 3(a^3+b^3+c^3)\geq (a^2+b^2+c^2)^2$
  2. Cauchy-Schwartz: $(1+1+1)(a^2+b^2+c^2)\geq (a+b+c)^2 \implies (a^2+b^2+c^2)\geq 3.$
  3. Combine two previous steps: $$ 3(a^3+b^3+c^3)\geq (a^2+b^2+c^2)^2\geq 3(a^2+b^2+c^2). $$ And therefore, this proves $(\star)$.

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