parmenides51 wrote: The real numbers $x,y,z, m, n$ are positive, such that $m + n \ge 2$. Prove that $x\sqrt{yz(x + my)(x + nz)} + y\sqrt{xz(y + mx)(y + nz)} + z\sqrt{xy(z + mx)(x + ny) }\le \frac{3(m + n)}{8} (x + y)(y + z)(z + x)$ By AM-GM $$\sum_{cyc}x\sqrt{yz(x + my)(x + nz)}\leq\frac{1}{2}\sum_{cyc}x(xy+nyz+xz+myz)=$$$$=\frac{1}{2}\sum_{cyc}(x^2y+x^2z+(m+n)xyz).$$Thus, it's enough to prove that $$(m+n)\sum_{cyc}(3x^2y+3x^2z+2xyz)\geq4\sum_{cyc}(x^2y+x^2z+(m+n)xyz)$$or $$(3m+3n-4)\sum_{cyc}(x^2y+x^2z)-6(m+n)xyz\geq0$$or $$(3m+3n-4)\sum_{cyc}(x^2y+x^2z-2xyz)+12(m+n-2)xyz\geq0.$$Done!