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If $a,b,c>0$ and $a+b+c=1$ prove inequality: $$\frac a{b^2 +c} + \frac b{c^2+a} + \frac c{a^2+b} \ge \frac 94$$

Shuhao Cao
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user133853
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3 Answers3

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since $$\sum_{cyc}\dfrac{a}{b+c^2}=\sum_{cyc}\dfrac{a^2}{ab+ac^2}$$ Use Cauchy-Schwarz inequality,we have $$\left(\sum_{cyc}\dfrac{a^2}{ab+ac^2}\right)\cdot\sum_{cyc}(ab+ac^2)\ge\left(\sum_{cyc}a\right)^2=1$$ so we only prove this following inequality $$\dfrac{1}{\sum_{cyc}(ab+ac^2)}\ge\dfrac{9}{4}$$ $$\Longleftrightarrow 4\ge9\sum_{cyc}ab+9\sum_{cyc}ac^2$$ $$\Longleftrightarrow 4(\sum_{cyc}a)^2\ge 9\sum_{cyc}ab+9\sum_{cyc}ac^2$$ $$\Longleftrightarrow 4\sum_{cyc}a^2\ge \sum_{cyc}ab+9\sum_{cyc}ac^2$$ since $$\sum_{cyc}a^2\ge \sum_{cyc}ab$$ so we only prove this $$3\sum_{cyc}a^2\ge 9\sum_{cyc}ac^2$$ $$\Longleftrightarrow \sum_{cyc}a^2\ge3\sum_{cyc}ac^2$$ $$\Longleftrightarrow \sum_{cyc}a\cdot\sum_{cyc}a^2\ge3\sum_{cyc}ac^2$$ $$\Longleftrightarrow \sum_{cyc}a^3+\sum_{cyc}ab^2\ge 2\sum_{cyc}ac^2$$ since Use AM-GM inequality $$\sum_{cyc}a^3+\sum_{cyc}ab^2=\sum_{cyc}(a^3+ab^2)\ge\sum_{cyc}(2a^2b)=2\sum_{cyc}ac^2$$ By done!

math110
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  • @math110 how is your argumentation before the first "since" ? – OBDA Mar 08 '14 at 10:55
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    @user133853: First, it is the policy of this site not to delete good content. Second, users have put effort into answering your question and deserve the credit they get. – robjohn Mar 12 '14 at 23:35
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Answer:

Multiplying a in the numerator and the denominator, you get

$$LHS = \frac{\sum a^{2}}{\sum (ac+b^{2}a)}$$ Applying the Cauchy-Schwarz inequality in the below steps

\begin{align} &\geqslant\frac{(\sum a)^{2}}{\sum ac + \sum b^{2}a}\\ &\geqslant \frac{1}{\sum ac + \frac{1}{3}\sum a \sum a^2} = \frac{1}{3\sum ac +\sum a (\sum a)^2}\\ &\geqslant \frac{3}{\sum ac + (\sum a)^{2}} \end{align}

Applying the inequality $$\sum ac \leqslant \frac{1}{3} (\sum a)^{2}$$

$$LHS \geqslant \frac{9}{4}$$

Edit: @9rm, Apply the Cauchy-Schwarz inequlality to $$\sum b^{2}a \leqslant (\sum b^4)^\frac{1}{2}\cdot (\sum a^2)^\frac{1}{2}.$$ This will reduce to $$\leqslant (abc)^\frac{1}{2}\cdot \sum a\cdot \sqrt{\frac{1}{3}}({\sum a})^{2}.$$ This will further reduce to $$\sqrt{\frac{1}{3}}\cdot \sum a\cdot \sqrt{\frac{1}{3}}\cdot {\sum a}^{2} = \frac{1}{3}\cdot \sum a(\sum a)^{2}.$$

Daniel Fischer
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Satish Ramanathan
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  • I did'nt get the step where you used $\sum a \sum a^2 \ge 3\sum b^2a$ .. can you please explain :) – r9m Mar 08 '14 at 01:19
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    The first step seems to be using the "identity" $$\sum_k\frac{a_k}{b_k}=\frac{\sum\limits_ka_k}{\sum\limits_kb_k}$$ – robjohn Mar 08 '14 at 02:34
  • @9rm as I understand it,that is wrong replacing $a=b=c=1/3$. Maybe it means something else,and that is why I have never liked this notation. – chubakueno Mar 08 '14 at 04:27
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By C-S $$\sum_{cyc}\frac{a}{b+c^2}=\sum_{cyc}\frac{a^2}{ab+ac^2}\geq\frac{(a+b+c)^2}{\sum\limits_{cyc}(ab+a^2b)}.$$ Thus, it remains to prove that $$4(a+b+c)^2\geq9(ab+ac+bc)+9(a^2b+b^2c+c^2a)$$ or $$4(a+b+c)^3\geq9(ab+ac+bc)(a+b+c)+9(a^2b+b^2c+c^2a)$$ or $$\sum_{cyc}(4a^3-6a^2b+3a^2c-abc)\geq0$$ and since by AM-GM $\sum\limits_{cyc}a^2c\geq3abc$, it remains to prove that $$\sum_{cyc}(2a^3-3a^2b+a^2c)\geq0$$ or $$\sum_{cyc}(2a^3-3a^2b+ab^2)\geq0$$ or $$\sum_{cyc}a(a-b)(2a-b)\geq0$$ or $$\sum_{cyc}\left(a(a-b)(2a-b)-\frac{1}{3}(a^3-b^3)\right)\geq0$$ or $$\sum_{cyc}(a-b)^2(5a+b)\geq0.$$ Done!

Michael Rozenberg
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