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Let $\{a,b,c\}\subset[1,2]$. Prove that: $$\frac{5}{3a+2b}+\frac{5}{3b+2c}+\frac{5}{3c+2a}\geq\frac{2}{a+b}+\frac{2}{b+c}+\frac{2}{c+a}$$

This inequality similar to the following. Let $\{a,b,c\}\subset[1,2]$. Prove that: $$\frac{3}{a+2b}+\frac{3}{b+2c}+\frac{3}{c+2a}\geq\frac{2}{a+b}+\frac{2}{b+c}+\frac{2}{c+a},$$ which is solvable: $$\sum_{cyc}\left(\frac{3}{a+2b}-\frac{2}{a+b}\right)=\sum_{cyc}\frac{a-b}{(a+2b)(a+b)}=$$ $$=\sum_{cyc}\left(\frac{a-b}{(a+2b)(a+b)}+\frac{1}{6}\left(\frac{1}{a}-\frac{1}{b}\right)\right)=\sum_{cyc}\frac{(a-b)^2(2b-a)}{6ab(a+2b)(a+b)}\geq0.$$

If we'll try to use this way, so we'll get $$\sum_{cyc}\left(\frac{5}{3a+2b}-\frac{2}{a+b}\right)=\sum_{cyc}\frac{b-a}{(3a+2b)(a+b)}=$$ $$=\sum_{cyc}\left(\frac{b-a}{(3a+2b)(a+b)}-\frac{1}{10}\left(\frac{1}{a}-\frac{1}{b}\right)\right)=\sum_{cyc}\frac{(a-b)^2(3a-2b)}{10ab(3a+2b)(a+b)},$$ which is nothing.

Also we can try the Ravi's substitution.

Let $a=y+z$, $b=x+z$ and $c=x+y$.

Hence, $x=\frac{b+c-a}{2}\geq\frac{1+1-2}{2}=0$, which says that $x$, $y$ and $z$ are non-negatives.

But if we wish to forget about $[1,2]$ so it's not happens

because this substitution gives a wrong inequality for $x=1$, $y=2$ and $z=0$.

Any hint would be desirable.

Thank you!

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  • $\begingroup$ it's true on $[0.666\dots,1]$ $\endgroup$
    – JMP
    Commented Mar 10, 2017 at 8:28
  • $\begingroup$ @JonMark Perry Yes of course! See my proof above. $\endgroup$ Commented Mar 10, 2017 at 9:22
  • $\begingroup$ Full expanding gives you need to prove $\sum_{cyc} -60 a^5 b^2 c + 60 a^5 b c^2 - 70 a^4 b^3 c + 170 a^4 b c^3 - 100 a^3 b^3 c^2 \ge 0$, but you will need something out of $\{a,b,c\}\subset[1,2]$... $\endgroup$
    – guest
    Commented Mar 10, 2017 at 10:59
  • $\begingroup$ @guest The term $\sum\limits_{cyc}(a^4c-a^4b)$ can be negative. $\endgroup$ Commented Mar 10, 2017 at 11:16

1 Answer 1

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We have that $\left\{ \begin{align} x = \dfrac{a + b}{2}\\ y = \dfrac{b + c}{2}\\ z = \dfrac{c + a}{2} \end{align} \right.$$\implies \left\{ \begin{align} a = z + x - y\\ b = x + y - z\\ c = y + z - x\end{align} \right. (x, y, z \in [1, 2])$. The original inequality becomes $$\frac{5}{5x - y + z} + \frac{5}{5y - z + x} + \frac{5}{5z - x + y} \ge \frac{1}{x} + \frac{1}{y} + \frac{1}{z}$$

$$\iff \left(\frac{5}{5x - y + z} - \frac{1}{x}\right) + \left(\frac{5}{5y - z + x} - \frac{1}{y}\right) + \left(\frac{5}{5z - x + y} - \frac{1}{z}\right) \ge 0$$

$$\iff \frac{y - z}{x(5x - y + z)} + \frac{z - x}{y(5y - z + x)} + \frac{x - y}{z(5z - x + y)} \ge 0$$

$$\iff \frac{y}{x(5x - y + z)} + \frac{z}{y(5y - z + x)} + \frac{x}{z(5z - x + y)} \ge \frac{z}{x(5x - y + z)} + \frac{x}{y(5y - z + x)} + \frac{y}{z(5z - x + y)}$$

Suppose that $x \ge y \ge z$.

Combined with $x, y, z \in [1, 2]$, the above inequality can be proven using rearrangement inequality.

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    $\begingroup$ You can't assume $x\ge y\geq z$ because the inequality is not symmetric. $\endgroup$ Commented Jun 28, 2019 at 6:21

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