Suppose that $a,b,c>0$. How to prove $$\frac{a}{7a+b}+\frac{b}{7b+c}+\frac{c}{7c+a}\le\frac38$$ ?

My first idea: By AM-GM, $$7a+b\geq \sqrt{7ab}$$ so $$\sum_{cyc} \frac{a}{7a+b}\le\sum_{cyc}\sqrt{\frac{a}{7b}}$$ but I am not sure if we can continue from here.

Also I try Cauchy-Schwarz: $$\sum_{cyc} \frac{a}{7a+b}\le\sqrt{a^2+b^2+c^2}\sqrt{\sum_{cyc} \frac{1}{(7a+b)^2}}.$$

Now what?

  • $\begingroup$ Clearly equality is achieved when $a=b=c$. If you call $x=b/a$, $y=c/b$ and $z=a/c$, you can change the problem into looking at the maximum of $$\frac{1}{7+x}+\frac{1}{7+y}+\frac{1}{7+z}$$ for $xyz=1$. Using standard Lagrange multipliers you find the unique maximum at $x=y=z=1$. $\endgroup$ – Crostul Dec 20 '19 at 23:05

By C-S $$\sum_{cyc}\frac{a}{7a+b}=\frac{3}{7}+\sum_{cyc}\left(\frac{a}{7a+b}-\frac{1}{7}\right)=\frac{3}{7}-\frac{1}{7}\sum_{cyc}\frac{b}{7a+b}=$$ $$=\frac{3}{7}-\frac{1}{7}\sum_{cyc}\frac{b^2}{7ab+b^2}\leq\frac{3}{7}-\frac{1}{7}\cdot\frac{(a+b+c)^2}{\sum\limits_{cyc}(7ab+b^2)}.$$ Id est, it's enough to prove that $$\frac{3}{7}-\frac{1}{7}\cdot\frac{(a+b+c)^2}{\sum\limits_{cyc}(7ab+b^2)}\leq\frac{3}{8}$$ or $$8(a+b+c)^2\geq3\sum\limits_{cyc}(7ab+a^2)$$ or $$\sum_{cyc}(a-b)^2\geq0$$ and we are done!


By AM-GM we have $$a^2b+ac^2+b^2c\geq3abc$$ and $$a^2c+ab^2+bc^2\geq 3abc$$ so that $$35(a^2b+ac^2+b^2c)+13(a^2c+ab^2+bc^2)\geq 3(35+13)abc=144abc.$$

Now, note that $$\frac38-\sum_{\text{cyc}} \frac{a}{7a+b}=\frac{35(a^2b+ac^2+b^2c)+13(a^2c+ab^2+bc^2)-144abc}{8 (7 a+b) (a+7 c) (7 b+c)},$$

which is non-negative by the previous result.

We have equality if and only if we have equality in both AM-GMs which implies $a=b=c$.

  • $\begingroup$ @Crostul Now with correct computations 😃 $\endgroup$ – Maximilian Janisch Dec 20 '19 at 23:09

Can be much detail, please


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    $\begingroup$ Can you ask it by comment under my post? Thank you! $\endgroup$ – Michael Rozenberg Dec 21 '19 at 16:41
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    $\begingroup$ I can't ask this while commenting on Your post as I don't have 50 reputation. $\endgroup$ – hpbhpb Dec 21 '19 at 18:37
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    $\begingroup$ OK. I used Cauchy-Schwarz inequality: for $b_i>0$ prove that: $\frac{a_1^2}{b_1}+\frac{a_2^2}{b_2}+...+\frac{a_n^2}{b_n}\geq\frac{(a_1+a_2+...+a_n)^2}{b_1+b_2+...+b_n}.$ $\endgroup$ – Michael Rozenberg Dec 21 '19 at 19:32
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    $\begingroup$ Thanks very much! $\endgroup$ – hpbhpb Dec 21 '19 at 20:38

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