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let $x,y,z$ are postive numbers,and such $$xy+zx+yz=1$$ find the maxum of $$\dfrac{1}{1+x^2}+\dfrac{1}{1+y^2}+\dfrac{z}{1+z^2}$$

my try

note $$x^2+1=x^2+xy+yz+xz=(x+y)(x+z)$$ $$y^2+1=(y+x)(y+z)$$ $$z^2+1=(z+x)(z+y)$$ $$\dfrac{1}{1+x^2}+\dfrac{1}{1+y^2}+\dfrac{z}{1+z^2}=\dfrac{y+z+x+z+z(x+y)}{(x+y)(y+z)(x+z)}$$ I think we must use $$8(a+b+c)(ab+bc+ac)\le 9(a+b)(b+c)(a+c)$$ $$LHS\le \dfrac{9}{8}\dfrac{x+y+z+z(x+y+1)}{x+y+z}=\dfrac{9}{8}(1+\dfrac{z(x+y+1)}{x+y+z})$$ then follow I can't works,Thank you

azimut
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2 Answers2

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Trigonometric substitution seems easier here. We may set $x = \tan\frac{A}{2}, y = \tan\frac{B}{2}, z = \tan\frac{C}{2}$, where $A, B, C$ are angles of a triangle, so that the constraint is satisfied. Then we have to find the maximum of $$f = \dfrac{1}{1+\tan^2\frac{A}{2}}+\dfrac{1}{1+\tan^2\frac{B}{2}}+\dfrac{\tan\frac{C}{2}}{1+\tan^2\frac{C}{2}} = \cos^2\dfrac{A}{2} + \cos^2\dfrac{B}{2}+ \dfrac{\sin C}{2}$$

Using $\cos \theta = 2 \cos^2 \frac{\theta}{2} - 1$, $$f = \dfrac{\cos A + 1}{2} + \dfrac{\cos B + 1}{2}+ \dfrac{\sin C }{2} = 1+\dfrac{\cos A + \cos B + \sin C}{2}$$

So we need to find the maximimum of $\cos A + \cos B + \sin C$, where $A, B, C$ are angles of a triangle. Now, $$\cos A + \cos B + \sin C = \cos A + \cos B + \sin (\pi - (A+B)) \\ = \frac{2}{\sqrt3}\left(\frac{\sqrt3}{2} \cos A + \frac{\sqrt3}{2} \cos B\right) + \frac{1}{\sqrt3}\left(\sqrt3 \sin A \cos B + \sqrt 3 \cos A \sin B\right)\\ \le \frac{1}{\sqrt3}\left(\frac{3}{4} + \cos^2 A + \frac{3}{4}+ \cos^2 B\right) + \frac{1}{2\sqrt3}\left(3 \sin^2 A +\cos^2 B + \cos^2 A +3\sin^2 B\right)\\ = \frac{\sqrt3}{2}+\frac{\sqrt3}{2}(\sin^2 A + \cos^2A) + \frac{\sqrt3}{2}(\sin^2 B + \cos^2 B) = \frac{3\sqrt 3}{2}$$

Hence $f \le 1 + \frac{3\sqrt3}{4}$, and equality is obtained when the AM-GMs we used have equality, i.e. $A = B = \frac{\pi}{6}, C = \frac{2\pi}{3}$ or $x = y = 2 - \sqrt{3}, z = \sqrt{3}$

Macavity
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  • Substituting $x=y=z=\sqrt{3}/3$ in the objective function, we obtain $9/4=2.25$. Subsituting $x = y = 2 - \sqrt{3}, z = \sqrt{3} $ in the objective function, we obtain $2, \left( 8-4,\sqrt {3} \right) ^{-1}+1/4=2.116025407 $. – user64494 Sep 29 '13 at 16:23
  • @user64494 Check again. On substituting $x=y=2-\sqrt3, z = \sqrt{3}$, I get $1+\dfrac{3\sqrt{3}}{4} \approx 2.299$. – Macavity Sep 29 '13 at 16:41
  • @ Macavity $$2, \left( \left( 2-\sqrt {3} \right) ^{2}+1 \right) ^{-1}+1/4=\frac {2}{2^2-2\cdot2\cdot\sqrt{3}+\sqrt{3}^2+1}= $$ $$ 2, \left( 8-4,\sqrt {3} \right) ^{-1}+1/4=2.116025407. $$ – user64494 Sep 29 '13 at 16:52
  • @user64494 $\dfrac{z}{1+z^2} =\dfrac{\sqrt{3}}{1+3}=\dfrac{\sqrt{3}}{4}\neq \dfrac{1}{4}$ – Macavity Sep 29 '13 at 16:56
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Due to symmetry of the formula, you can suppose x=y=t. Hence $$t^2+2tz=1$$ $$z=\frac{1-t^2}{2t}$$

$$f(t)=\frac{2}{1+t^2}+\frac{2t(1-t^2)}{4t^2+(1-t^2)^2}=2\frac{1+t+t^2-t^3}{(1+t^2)^2}$$

$$f'(t)=2\frac{(-3t^2+2t+1)(1+t^2)-4t(-t^3+t^2+t+1)}{(1+t^2)^3}$$

$$f'(t)=2\frac{t^4-2t^3-6t^2-2t+1}{(1+t^2)^3}$$

$P(t)=t^4-2t^3-6t^2-2t+1=(t+1)(t^3-3t^2-3t+1)=(t+1)^2(t^2-4t+1)$

Hence there are extremum in $t=-1$ (not really as it is a double root), and $t={2\pm\sqrt{3}}$. You can easily verify the maximum is for $2-\sqrt{3}$.

Xoff
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