6
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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
4
$\begingroup$

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!

$\endgroup$
3
$\begingroup$

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 $a=b=c$.

$\endgroup$
  • $\begingroup$ @Crostul Now with correct computations 😃 $\endgroup$ – Maximilian Janisch Dec 20 '19 at 23:09
0
$\begingroup$

Can be much detail, please

$$\frac{3}{7}-\frac{1}{7}\sum_{cyc}\frac{b^2}{7ab+b^2}\leq\frac{3}{7}-\frac{1}{7}\cdot\frac{(a+b+c)^2}{\sum\limits_{cyc}(7ab+b^2)}.$$

$\endgroup$
  • 4
    $\begingroup$ Can you ask it by comment under my post? Thank you! $\endgroup$ – Michael Rozenberg Dec 21 '19 at 16:41
  • 1
    $\begingroup$ 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
  • 2
    $\begingroup$ 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
  • 3
    $\begingroup$ Thanks very much! $\endgroup$ – hpbhpb Dec 21 '19 at 20:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.