socrates wrote: Let $a, b, c$ be the lengths of the sides of a triangle, and $h_a, h_b, h_c$ the lengths of the heights corresponding to the sides $a, b, c,$ respectively. If $t \geq \frac{1} {2}$ is a real number, show that there is a triangle with sidelengths $$ t\cdot a + h_a, \ t\cdot b + h_b , \ t\cdot c + h_c.$$ $\frac{1}{2}a+h_a,\frac{1}{2}b+h_b,\frac{1}{2}c+h_c$ just are sides of $\triangle A_1B_1C_1$, $\triangle A_1B_1C_1$ area is $S_1$,then $S_1=(1+\frac{\sqrt{3}}{2})S$ But prove it i very hard. BQ

It is enough to prove the case $t=\frac{1}{2}.$ Using trigonometry, we have to prove that $$\sin A(1-2\sin B-2 \sin C)<\sin B+\sin C-2\sin B \sin C,$$which is obvious using $\sin A<\sin B+\sin C.$ Also, when $A=0^+, B=C=90^-$ we have $t\geq \frac{1}{2},$ so this is the best constant.