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how compute $\max\{x+z\}$ and $\max\{1+y^2\}$? such that $x$,$y$,$z$ satisfied $$\begin{cases} xy+xz+yz=1 & \\ x\ge0 \\ y\ge0\\ z\ge0\\ \end{cases} $$ i face with this problem when i try solve here

Thanks in advance

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M.H
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2 Answers2

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We can try Lagrange Multiplier

Alternatively, we know if $A+B+C=\frac\pi2, \sum \tan A\tan B=1$

As $x,y,z\ge0,$ we can choose $A,B,C\in [0,\frac\pi2]$

Now, $1+y^2=1+\tan^2C=\sec^2C$ which tends to $\infty$ as $C\to\infty$

$x+z=\tan A+\tan C$ which tends to $\infty$ as $A$ or $C\to\infty$

lab bhattacharjee
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  • i edit my question please see once again my question – M.H Jun 09 '13 at 13:52
  • @MaisamHedyelloo, looking into it – lab bhattacharjee Jun 09 '13 at 13:58
  • if if $x+z$ tend to $\infty$ how do u justify minimum value of $\frac{1}{\sqrt{1+y^2}}+2\frac{1}{\sqrt{y+1}\sqrt{x+z}}$ in this question http://math.stackexchange.com/questions/413683/show-that-frac-1-sqrtxy-frac-1-sqrtyz-frac-1-sqrtzx-geq2 – M.H Jun 09 '13 at 14:05
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$(x+z)$ is unbounded above.Reason: Let $M>0$ be given and let $y=0,z=M,x=\frac{1}{M}$ Then $(x+z)>M$ so it is unbounded above.

$(1+y^2)$ is also unbounded. Reason take $y=M,z=0,x=\frac{1}{M}$ then $(1+y^2)>M$

Abhra Abir Kundu
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