Let $a$, $b$ and $c$ be positive numbers such that $a+b+c=3$. Prove that: $$\frac{a}{\sqrt{b^2+c^2}}+\frac{b}{\sqrt{a^2+c^2}}+\frac{c}{\sqrt{a^2+b^2}}\geq\frac{1}{\sqrt{a^2+bc}}+\frac{1}{\sqrt{b^2+ac}}+\frac{1}{\sqrt{c^2+ab}}$$

I tried to use C-S, Holder, SOS, Rearrangement and more, but without some success.

  • $\begingroup$ Is it possible to prove the inequality with 1 instead of a,b,c of each numerator of the 3 terms in the LHS? $\endgroup$ – Raizen Oct 31 '16 at 11:36
  • $\begingroup$ @Raizen It would wrong. Try $c\rightarrow0^+$ and $a=b$ $\endgroup$ – Michael Rozenberg Oct 31 '16 at 13:08
  • $\begingroup$ @MichaelRozenberg $a=b, c\to 0^+$ gives equality, which seems to suggest Schur is at work. $\endgroup$ – Macavity Oct 31 '16 at 16:53
  • $\begingroup$ This is another case of a general set: if positive $a++c=3$, and $1\leq k \leq 5$ then $$\sum_{cyc}\frac{a}{\sqrt{kb^2+c^2}}\geq \sum_{cyc}\frac{1}{\sqrt{ka^2+bc}}$$ Here the $k=1$ end is sharp (for $k<1$ the inequality fails); but the $k=5$ end actually goes to a bit more than $5.15$. $\endgroup$ – Mark Fischler Apr 3 '17 at 23:52
  • $\begingroup$ @Michael Rozenberg Do you know the kulp quartic ? the cartesian equation is : $$y=\frac{a}{\sqrt{a^2+x^2}}$$ or if we make a good substitution on $y$ we have : $$y'=\frac{b}{\sqrt{a^2+x^2}}$$.Maybe it would be helpfull. $\endgroup$ – user448747 Jun 30 '17 at 15:28

We begin with a first substitution we put :




The initial inequality become :

$$\frac{v-u}{\sqrt{(\frac{v+u}{2})^2+(\frac{2w-v-u}{2})^2}}+\frac{\frac{v+u}{2}}{\sqrt{(v-u)^2+(\frac{2w-v-u}{2})^2}}+\frac{\frac{2w-v-u}{2}}{\sqrt{(v-u)^2+(\frac{u+v}{2})^2}}\geq \frac{1}{\sqrt{(v-u)^2+(\frac{2w-v-u}{2})(\frac{v+u}{2})}}+\frac{1}{\sqrt{(\frac{v+u}{2})^2+(\frac{2w-v-u}{2})(v-u)}}\frac{1}{\sqrt{(\frac{2w-v-u}{2})^2+(v-u)(\frac{v+u}{2})}}$$

We get $w+v-u=3$

Now we make a second substitution :




We get this :


The initial inequality become

$$\frac{2\frac{r-p}{2}}{\sqrt{(\frac{r+p}{2})^2+(\frac{2q-r-p}{2})^2}}+\frac{\frac{r+p}{2}}{\sqrt{(2\frac{r-p}{2})^2+(\frac{2q-r-p}{2})^2}}+\frac{\frac{2q-r-p}{2}}{\sqrt{(2\frac{r-p}{2})^2+(\frac{r+p}{2})^2}}\geq \frac{\sqrt{r^2+q^2-p^2}}{\sqrt{(2\frac{r-p}{2})^2+(\frac{2q-r-p}{2})(\frac{r+p}{2})}}+\frac{\sqrt{r^2+q^2-p^2}}{\sqrt{(\frac{r+p}{2})^2+(\frac{2q-r-p}{2})(2\frac{r-p}{2})}}\frac{\sqrt{r^2+q^2-p^2}}{\sqrt{(\frac{2q-r-p}{2})^2+(2\frac{r-p}{2})(\frac{r+p}{2})}}$$

We make a last substitution :




With $P\geq 1$ if we put $p\geq q \geq r$

So the initial condition become :


Wich is equivalent to :


The initial inequality become :

$$\frac{2\frac{1-P}{2}}{\sqrt{(\frac{1+P}{2})^2+(\frac{2Q-1-P}{2})^2}}+\frac{\frac{1+P}{2}}{\sqrt{(2\frac{1-P}{2})^2+(\frac{2Q-1-P}{2})^2}}+\frac{\frac{2Q-1-P}{2}}{\sqrt{(2\frac{1-P}{2})^2+(\frac{1+P}{2})^2}}\geq \frac{\sqrt{1+Q^2-P^2}}{\sqrt{(2\frac{1-P}{2})^2+(\frac{2Q-1-P}{2})(\frac{1+P}{2})}}+\frac{\sqrt{1+Q^2-P^2}}{\sqrt{(\frac{1+P}{2})^2+(\frac{2Q-1-P}{2})(2\frac{1-P}{2})}}\frac{\sqrt{1+Q^2-P^2}}{\sqrt{(\frac{2Q-1-P}{2})^2+(2\frac{1-P}{2})(\frac{1+P}{2})}}$$

So to conclude it sufficient to combine the condition with the inequality to obtain an inequality with one variable .

  • $\begingroup$ I see two variables. How can you make an inequality with one variable? $\endgroup$ – Michael Rozenberg Jul 12 '17 at 10:20
  • $\begingroup$ If you replace the value of $Q$ in the inequality you have an inequality just with $P$ $\endgroup$ – max8128 Jul 12 '17 at 14:53
  • $\begingroup$ Why $w+v-u=3$? I think $w+v-u=a+2b+c$. $\endgroup$ – Michael Rozenberg Jul 12 '17 at 15:06
  • $\begingroup$ Sorry I'm going to correct it . $\endgroup$ – max8128 Jul 12 '17 at 15:16
  • $\begingroup$ Now it would be correct... $\endgroup$ – max8128 Jul 12 '17 at 15:30

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.