2
$\begingroup$

If $a$,$b$ and $c$ positive real numbers such that $a+b+c=1$, prove $\dfrac{b^2}{a+b^2}+\dfrac{c^2}{b+c^2}+\dfrac{a^2}{c+a^2} \geqslant \dfrac{3}{4}$. I have tried several methods to solve this,but can't get any result. Any idea?

$\endgroup$
  • 1
    $\begingroup$ i think $a,b,c$ are positive $\endgroup$ – Dr. Sonnhard Graubner Jan 6 '16 at 18:08
  • $\begingroup$ of course,i will change task now $\endgroup$ – chaos Jan 6 '16 at 18:09
  • $\begingroup$ from where does this come? $\endgroup$ – Dr. Sonnhard Graubner Jan 6 '16 at 18:22
  • $\begingroup$ If it is helpful, $\frac{3}{4}$ is attained where $a=b=c$. $\endgroup$ – CommonerG Jan 6 '16 at 20:57
3
$\begingroup$

By C-S and Vasc we obtain $\sum\limits_{cyc}\frac{a^2}{a^2+c}=\sum\limits_{cyc}\frac{a^4}{a^4+a^2c(a+b+c)}\geq\frac{(a^2+b^2+c^2)^2}{\sum\limits_{cyc}(a^4+a^3c+a^2b^2+a^2bc)}\geq$

$\geq\frac{(a^2+b^2+c^2)^2}{\sum\limits_{cyc}(a^4+a^2b^2+a^2bc)+\frac{(a^2+b^2+c^2)^2}{3}}=\frac{3(a^2+b^2+c^2)^2}{\sum\limits_{cyc}(4a^4+5a^2b^2+3a^2bc)}$.

Id est, it remains to prove that $4(a^2+b^2+c^2)^2\geq\sum\limits_{cyc}(4a^4+5a^2b^2+3a^2bc)$, which is $\sum\limits_{cyc}c^2(a-b)^2\geq0$. Done!

$\endgroup$
  • $\begingroup$ It is a good answer, but it is quite cryptical. I bet C-S is Cauchy-Schwarz, but what it Vasc? To improve the rendering would be a good idea, too, IMHO. $\endgroup$ – Jack D'Aurizio Jan 6 '16 at 20:01
  • $\begingroup$ The Vasc;'s inequality it's the following. $(a^2+b^2+c^2)^2\geq3(a^3c+b^3a+c^3b)$. It's true because $(a^2+b^2+c^2)^2-3(a^3c+b^3a+c^3b)=\frac{1}{2}\sum\limits_{cyc}(a^2-b^2+ab-2ac+bc)^2\geq0$. $\endgroup$ – Michael Rozenberg Jan 6 '16 at 20:12

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.