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Let $a,b,c>0; a+b+c=1$. Prove the inequality $$\frac a{a+b^2}+\frac b{b+c^2}+\frac c{c+a^2}\le\frac14\left(\frac1a+\frac1b+\frac1c\right)$$

My work so far:

I tried AM-GM and used fact $a+b+c=1$.

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  • $\begingroup$ What was the result of your work so far? $\endgroup$
    – abiessu
    Commented Sep 20, 2016 at 17:49
  • $\begingroup$ I would have written, if I considered them important $\endgroup$
    – Roman83
    Commented Sep 20, 2016 at 18:02

3 Answers 3

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An homogenization and a full expanding give: $$\sum\limits_{cyc}(a^6b^2+2a^5b^3+a^5c^3+2a^4b^4+a^6bc-2a^5b^2c+4a^5c^2b+$$ $$+a^4b^3c+4a^4c^3b-7a^4b^2c^2-7a^3b^3c^2)\geq0$$ which is obvious by AM-GM.

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  • $\begingroup$ How do you apply AM-GM ? Perhaps you're adding two different applications of AM-GM ? $\endgroup$ Commented Sep 20, 2016 at 18:46
  • $\begingroup$ @Ewan Delanoy 5 For example, by AM-GM $a^6bc+a^4b^3c\geq2a^5b^2c$ $\endgroup$ Commented Sep 20, 2016 at 18:53
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Also we can us C-S and $uvw$:

We need to prove that $$\sum\limits_{cyc}\left(\frac{a}{a+b^2}-1\right)\leq\frac{ab+ac+bc}{4abc}-\frac{3}{a+b+c}$$ or $$\sum\limits_{cyc}\frac{b^2}{a+b^2}+\frac{ab+ac+bc}{4abc}-\frac{3}{a+b+c}\geq0$$ Now by C-S: $$\sum\limits_{cyc}\frac{b^2}{a+b^2}\geq\frac{(a+b+c)^2}{a+b+c+a^2+b^2+c^2}=\frac{a+b+c}{(a+b+c)^2+a^2+b^2+c^2}$$ Id est, it remains to prove that $$\frac{a+b+c}{(a+b+c)^2+a^2+b^2+c^2}+\frac{ab+ac+bc}{4abc}-\frac{3}{a+b+c}\geq0$$ Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.

Hence, we need to prove that $f(w^3)\geq0$, where $f$ is a decreasing function,

which says that it's enough to prove $f(w^3)\geq0$ for a maximal value of $w^3$.

Since $a$, $b$ and $c$ are positive roots of the equation $$(x-a)(x-b)(x-c)=0$$ or $$w^3=x^3-3ux^2+3v^2x$$

we see that $w^3$ gets a maximal value for equality case of two variables

and since $f(w^3)\geq0$ is an homogeneous inequality, it's enough to prove it for $b=c=1$, which gives $$(a-1)^2(2a^2+3a+6)\geq0$$ Done!

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$$\frac{a}{a+b^2}= \frac{a}{a^2+ab+ac+b^2} = \frac{a}{(a^2+ac)+(ab+b^2)} \le \frac{a}{4} \cdot \left( \frac{1}{a^2+ac}+\frac{1}{ab+b^2} \right)=$$

$$= \frac{1}{4}\cdot \left( \frac{1}{a+c}+\frac{1}{b}-\frac{1}{a+b} \right)$$

$$\frac{a}{a+b^2}+\frac{b}{b+c^2}+\frac{c}{c+a^2} \le \frac{1}{4}\cdot \left( \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \right)$$

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