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Consider three real numbers $a \ge b \ge c \gt 0$. If $(a^x-b^x-c^x)(x-2) \gt 0$ for any rational number $x \neq 2$, show that

  1. $a, b, c$ can be the length of the three sides of a triangle $ABC$;
  2. $ABC$ is a right-angled triangle.

My try: Since, for $x\neq 2, (a^x-b^x-c^x)(x-2) \gt 0$ either $(a^x-b^x-c^x)\gt 0$ and $(x-2) \gt 0$ or $(a^x-b^x-c^x) \lt 0$ and $(x-2) \lt 0$.If $(x-2) \lt 0$ then $(a^x-b^x-c^x)\lt 0$ and for $x=1 , a \lt b+c$. Hence $a, b, c $ be the length of the side of the triangle $ABC$. But cannot proceed further. Please help me. Thanks in advance.

Debprasad Kundu
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    $f(x)=a^x-b^x-c^x$ is a continuous function of $x$ for $x$ real: it it the sum of three continuous functions. If $f(x)>0$ for $x>2$ and $f(x)<0$ for $x<2$ for rational $x$ then the same is true for all real $x$, so $f(2)=0$ and $a^2=b^2+c^2$ – Henry Jul 20 '21 at 07:52
  • Thank you. For first part I take for particular x=1. How to prove it in general – Debprasad Kundu Jul 20 '21 at 08:02

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