Maximize $P=\frac{1}{\sqrt{1+x^{2}}}+\frac{1}{\sqrt{1+y^{2}}}+\frac{1}{\sqrt{1+z^{2}}}$

Question

For $x,y,z$ are positive real numbers that satisfy $xy+yz+xz=1$. Maximize $$P=\frac{1}{\sqrt{1+x^{2}}}+\frac{1}{\sqrt{1+y^{2}}}+\frac{1}{\sqrt{1+z^{2}}}.$$

I think if we let $x=\tan A;y=\tan B;z=\tan C$, then

$$P\Leftrightarrow \displaystyle \text {sin}A+\text {sin}B+ \text{sin}C\leq \frac{3}{2}. \hspace{2cm}(1)$$

But I can't prove $(1)$.

$P$ and your constraint is symmetric with interchange of $x \leftrightarrow y \leftrightarrow z$. So... assuming the final solution is also so symmetric, what simplification can you make? — David G. Stork, Mar 19 '17 at 05:07
@David. Your comment that P(x,y,z) and the constraint are invariant under any permutation of (x,y,z) seems to be a strong argument for $ x_m = y_m = z_m$. ( m stands for maximum) I cannot prove it, once this is established the problem is solved. Can anyone prove it, or am I wrong altogether? — Peter Szilas, Mar 20 '17 at 09:49
@PeterSzilas: You can use Lie group theory and invariants to prove that if the problem respects an invariance, the solution will as well. Practically, find the solution for $x = y = z$ and then verify that it is optimum, as Michael Rozenberg does, below. — David G. Stork, Mar 20 '17 at 16:03

score 1 · Answer 1 · answered Mar 19 '17 at 05:08

1

Hint : You're making a mistake.

You have to prove this $$ \cos A + \cos B + \cos C \le \frac{3}{2} $$ because : $$ 1+\tan^2 \theta = \sec^2 \theta $$

answered Mar 19 '17 at 05:08

Vishnu V.S

975

Okay but now how will you prove this? – Archis Welankar Mar 19 '17 at 05:10
$\cos C = \cos (A+B)$ Try this. – Vishnu V.S Mar 19 '17 at 05:11
Wait. Sorry. It's $\cos C = -\cos (A+B)$ – Vishnu V.S Mar 19 '17 at 05:12
The suggested transformations $x=\tan A$, etc. do not work here. The substitutions are $x=\cot A$, etc. – Bob Jones Mar 19 '17 at 05:15
Where is it given that $A,B,C$ are angles of a triangle? – Archis Welankar Mar 19 '17 at 05:16
You made them the angles of a triangle when you set $$\tan A \tan B + \tan B \tan C + \tan A \tan C = 1$$. This is true only if $$A+B+C=\pi$$. – Vishnu V.S Mar 19 '17 at 05:19
@BobJones Those transformations do not work here because they do not satisfy : $xy+yz+xz=1$ – Vishnu V.S Mar 19 '17 at 05:23
What? If $A=B=C=\frac{\pi}{3}$, does $\tan A$, etc. satisfy that equality? Try it. The identity $\tan A$, etc. satisfy is $xyz=x+y+z$. – Bob Jones Mar 19 '17 at 05:38
@BobJones Oh... my bad. Yeah. They satisfy :$$\tan \frac{A}{2}\tan \frac{B}{2}+\tan \frac{B}{2}\tan \frac{C}{2}+\tan \frac{A}{2}\tan \frac{C}{2}=1$$. Yeah. So the $\cot $ transformation ought to work. – Vishnu V.S Mar 19 '17 at 05:45

score 1 · Accepted Answer · answered Mar 19 '17 at 05:20

1

For $x=y=z=\frac{1}{\sqrt3}$ we get a value $\frac{3\sqrt3}{2}$.

We'll prove that it's a maximal value.

Indeed, we need to prove that $$\sum_{cyc}\frac{1}{\sqrt{x^2+xy+xz+yz}}\leq\frac{3\sqrt3}{2\sqrt{xy+xz+yz}}$$ or $$\sum_{cyc}\sqrt{x+y}\leq\frac{3\sqrt{3(x+y)(x+z)(y+z)}}{2\sqrt{xy+xz+yz}}.$$ But by C-S $$\left(\sum_{cyc}\sqrt{x+y}\right)^2\leq(1+1+1)\sum_{cyc}(x+y)=6(x+y+z).$$

Thus, it remains to prove that $$\sqrt{6(x+y+z)}\leq\frac{3\sqrt{3(x+y)(x+z)(y+z)}}{2\sqrt{xy+xz+yz}}$$ or $$9(x+y)(x+z)(y+z)\geq8(x+y+z)(xy+xz+yz)$$ or $$\sum_{cyc}z(x-y)^2\geq0.$$ Done!

answered Mar 19 '17 at 05:20

Michael Rozenberg

194,933

1

Very nice solution. I checked the following $$\frac{1}{\sqrt{x^2+1}}+\frac{1}{\sqrt{y^2+1}}+\frac{1}{\sqrt{y^2+1}}\ge 2,$$and it is true. – TATA box Jun 16 '23 at 03:01
Yes, it's true and easy. – Michael Rozenberg Jun 16 '23 at 03:10
Did you use Holder ? – TATA box Jun 16 '23 at 03:23
No. Squaring and C-S. – Michael Rozenberg Jun 16 '23 at 05:06
1

The following is also true $$\frac{3\sqrt{15}}{5}\ge \frac{1}{\sqrt{2a^2+1}}+\frac{1}{\sqrt{2b^2+1}}+\frac{1}{\sqrt{2c^2+1}}\ge 2.$$ – TATA box Jun 17 '23 at 04:51

score 1 · Answer 3 · edited Apr 13 '17 at 12:20

1

Like How to show that the triangle is equilateral triangle?,

$\cot A\cot B+\cot B\cot C+\cot C\cot A=1$ with $A+B+C=\pi$

Now use this

edited Apr 13 '17 at 12:20

Community

1

answered Mar 19 '17 at 06:22

lab bhattacharjee

274,582

Maximize $P=\frac{1}{\sqrt{1+x^{2}}}+\frac{1}{\sqrt{1+y^{2}}}+\frac{1}{\sqrt{1+z^{2}}}$

3 Answers3