Skip to main content
MathOverflow

Return to Answer

Bounty Ended with 50 reputation awarded by Loreno Heer
deleted 3 characters in body
Source Link
Fedor Petrov
Fedor Petrov
  • 108.8k
  • 9
  • 264
  • 459

I am afraid it is false. Take $F=0$, $A=C$, $E=2A$, $\mu=1/2$. Then we are given $\cos B+\cos D=2\cos A$ and should prove $\cos B/2+\cos D/2\geqslant 2\cos A/2$. But if $\cos B=x$, then $\cos x/2=\sqrt{(1+x)/2}$$\cos B/2=\sqrt{(1+x)/2}$, this function is concave, thus inverse inequality    $\cos B/2+\cos D/2\leqslant 2\cos A/2$ holds holds.

I am afraid it is false. Take $F=0$, $A=C$, $E=2A$, $\mu=1/2$. Then we are given $\cos B+\cos D=2\cos A$ and should prove $\cos B/2+\cos D/2\geqslant 2\cos A/2$. But if $\cos B=x$, then $\cos x/2=\sqrt{(1+x)/2}$, this function is concave, thus inverse inequality  $\cos B/2+\cos D/2\leqslant 2\cos A/2$ holds.

I am afraid it is false. Take $F=0$, $A=C$, $E=2A$, $\mu=1/2$. Then we are given $\cos B+\cos D=2\cos A$ and should prove $\cos B/2+\cos D/2\geqslant 2\cos A/2$. But if $\cos B=x$, then $\cos B/2=\sqrt{(1+x)/2}$, this function is concave, thus inverse inequality  $\cos B/2+\cos D/2\leqslant 2\cos A/2$ holds.

Source Link
Fedor Petrov
Fedor Petrov
  • 108.8k
  • 9
  • 264
  • 459

I am afraid it is false. Take $F=0$, $A=C$, $E=2A$, $\mu=1/2$. Then we are given $\cos B+\cos D=2\cos A$ and should prove $\cos B/2+\cos D/2\geqslant 2\cos A/2$. But if $\cos B=x$, then $\cos x/2=\sqrt{(1+x)/2}$, this function is concave, thus inverse inequality $\cos B/2+\cos D/2\leqslant 2\cos A/2$ holds.