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Let $a,b,c\ge 0$ such that $a+b+c\le 1$, prove that $$3(a+b+c)-(a^2+b^2+c^2-ab-bc-ac)\ge (\sqrt{a}+\sqrt{b}+\sqrt{c})^2\tag{1}$$

I conjecture: Let $a_{i}\ge 0$, $i=1,2,\cdots$, $a_{1}+a_{2}+\cdots+a_{n}\le 1$, $n\ge 3$,then $$n(a_{1}+a_{2}+\cdots+a_{n})-(a^2_{1}+a^2_{2}+\cdots+a^2_{n}-a_{1}a_{2}-a_{2}a_{3}-\cdots-a_{n}a_{1})\ge (\sqrt{a_{1}}+\sqrt{a_{2}}+\cdots+\sqrt{a_{n}})^2$$ This inequality is stronger than Cauchy-Schwarz inequality,because $$(a^2+b^2+c^2-ab-bc-ac)\ge 0$$ Applying Cauchy-Schwarz inequality $$3(a+b+c)\ge(\sqrt{a}+\sqrt{b}+\sqrt{c})^2$$ How to prove the required statement (1)?

Did
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  • Maybe, if you look at the proof of CS-inequality in the case of $(1, 1, 1), (\sqrt a, \sqrt b, \sqrt c)$, you can get some hint to the margin by which the inequality is true. If you could compare that to $a^2 + b^2 + c^2 -ab-bc-ac$, you would be done. – Arthur May 08 '15 at 06:56

2 Answers2

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Since $a+b \leqslant 1 $ by Cauchy - Schwarz we get $$\sqrt{a} +\sqrt{b} \leq \sqrt{2}\cdot \sqrt{a+b} \leqslant \sqrt{2} $$ hence $$\frac{(\sqrt{a} +\sqrt{b})^2}{2} \leqslant 1$$ multiplying both sides of the above inequality by $(\sqrt{a} -\sqrt{b} )^2 $ we get $$\frac{(a-b)^2 }{2} \leqslant (\sqrt{a} -\sqrt{b})^2 . $$

Now we have

\begin{align} 3 (a+b+c ) -(\sqrt{a} +\sqrt{b} +\sqrt{c})^2 &= 2a +2b +2c -2\sqrt{ab} -2\sqrt{ac} -2\sqrt{bc} \\ &= (\sqrt{a} -\sqrt{b} )^2 +(\sqrt{a} -\sqrt{c} )^2 +(\sqrt{c} -\sqrt{b} )^2\\ &\geqslant \frac{1}{2} \cdot \left ( (a-b)^2 + (a-c)^2 +(c-b)^2 \right)\\ &= a^2 +b^2 +c^2 -ab -ac -ab \end{align}

what completes the proof.

Daniel Fischer
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  • It's Nice+!,so for $n$,we only $n-1\ge 2$,then use your methods to solve it –  May 08 '15 at 08:37
  • The conjecture is not true for $n>3 .$ For example if you take $n=4,$ and $a_1 =\frac{1}{4} , a_2 =\frac{1}{9} , a_3 =\frac{1}{16} , a_4 =\frac{1}{25} $ then the inequality you conjectured is not true for these numbers. –  May 08 '15 at 08:56
  • Oh,If add $n$ is odd is true? –  May 08 '15 at 10:18
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We need to prove that $$\sum_{cyc}(3a-a^2+ab)\geq\sum_{cyc}(a+2\sqrt{ab})$$ or $$2\sum_{cyc}(a-\sqrt{ab})\geq\sum_{cyc}(a^2-ab)$$ or $$2\sum_{cyc}\left(\sqrt{a}-\sqrt{b}\right)^2\geq\sum_{cyc}(a-b)^2$$ or $$\sum_{cyc}(\sqrt{a}-\sqrt{b})^2\left(2-(\sqrt{a}+\sqrt{b})^2\right)\geq0,$$ for which it's enough to prove that $$\sum_{cyc}(\sqrt{a}-\sqrt{b})^2\left(2(a+b+c)-(\sqrt{a}+\sqrt{b})^2\right)\geq0$$ or

$$\sum_{cyc}(\sqrt{a}-\sqrt{b})^2\left((\sqrt{a}-\sqrt{b})^2+2c\right)\geq0,$$ which is obvious.

Done!

Michael Rozenberg
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