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Let $f_1,f_2$ be two real valued functions defined on the real line.Define two functions $g,h$ by $g(x)=max${$f_1(x),f_2(x)$} and $h(x)=min${$f_1(x),f_2(x)$}.Then $g(x^2)+h(x)^2+3g(x)h(x)=f_1(x)^2+f_2(x)^2+3f_1(x)f_2(x)$ holds for all $x\in \mathbb R$

$1.$Always

$2.$ only of $f_1(x)=f_2(x)$ for all $x\in \mathbb R$

$3.$only $f_1$ and $f_2$ are both positive functions or both negative functions.

$4.$ only if atleast one of the functions $f_1$ and $f_2$ is identically zero

Solution:

if we choose $f_1(x)=1,f_2(x)=2$,then $g(x^2)+h(x)^2+3g(x)h(x)=2+1^2+3(2)(1)=9$

and

$f_1(x)^2+f_2(x)^2+3f_1(x)f_2(x)=1^2+2^2+3(2)(1)=11$,

So,

$g(x^2)+h(x)^2+3g(x)h(x)\neq f_1(x)^2+f_2(x)^2+3f_1(x)f_2(x)$

Hence,option $1,3,4$ get discarded so,the only remaining option i.e option 2.

must be true.

But, it is given that option $1$ is true.

Please point out my mistake

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    $\begingroup$ may be, it should be hold if the rhs is $f_1(x^2)$ and $f_2(x^2)$ instead of $f_{1}^2(x)$ and $f_{2}^2(x)$. And you can always prove the statement if you remain mot sure. $\endgroup$ – Reza Habibi Jun 9 '18 at 9:27
  • $\begingroup$ @:Reza Habibi:there is no tyo in the post,i wanted to know what is "wrong" with my counterexample? $\endgroup$ – P.Styles Jun 9 '18 at 9:38
  • $\begingroup$ I suspect that it's supposed to be $(g(x))^2+(h(x))^2+3g(x)h(x)$. The moment you start to much about with the argument to one of the functions but not the other, then things usually become very unequal very quickly. $\endgroup$ – Arthur Jun 9 '18 at 10:38
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Hint, assuming my comment above is true: Note that $f_1(x)+f_2(x)=g(x)+h(x)$ and $f_1(x)f_2(x)=g(x)h(x)$.

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