parmenides51 wrote: a) Prove that for all real numbers $k,l,m$ holds : $$(k+l+m)^2 \ge 3 (kl+lm+mk)$$When does equality holds? b) If $x,y,z$ are positive real numbers and $a,b$ real numbers such that $$a(x+y+z)=b(xy+yz+zx)=xyz,$$prove that $a \ge 3b^2$. When does equality holds? First is well-known. For second use that $(xy+yz+zx)^2 \geq 3xyz(x+y+z)$ for getting our result.

Part b is remarkably similar to 2014 British MO Round 2 Problem 2, so I will use the same trick I used to solve that problem. $a(x+y+z)=b(xy+yz+zx)=bxyz(1/x+1/y+1/z)=b^2(xy+yz+zx)(1/x+1/y+1/z)\geq3b^2(x+y+z)$ where the inequality at the end is true if and only if $xy/z + yz/x + zx/y \geq x+y+z$ which is just Muirhead or AMGM.

a) ${{(k+l+m)}^{2}}\ge 3(kl+lm+mk)\Leftrightarrow {{(k-l)}^{2}}+{{(l-m)}^{2}}+{{(m-k)}^{2}}\ge 0$

b) ${{(xy+yz+zx)}^{2}}\ge 3xyz(x+y+z)\Leftrightarrow {{(xy-yz)}^{2}}+{{(yz-zx)}^{2}}+{{(zx-xy)}^{2}}\ge 0$