ta có:x2+xy+xz=3yz⇔(x+y)(y+z)=4yzđặtx+y=a;z+x=b⇒(a−b)2=(y−z)2,ab=4yzlại có$:a^3+b^3=(a+b)(a^2-ab+b)^2\leq \sqrt{2(a^2+b^2)}[(a-b)^2+ab]=\sqrt{2[(z+y)^2+4yz]}(y+z)^2\leq \sqrt{4(y+z)^2}(y+z)^2=2(y+z)^2(1)tiếp:3(x+y)(y+z)(z+x)=12yz(y+z)\leq3(y+z)^2(y+z)=3(y+z)^3(2)$$(1)+(2)=đpcm$
ta có
:x2+xy+xz=3yz⇔(x+y)(y+z)=4yzđặt
x+y=a;z+x=b⇒(a−b)2=(y−z)2,ab=4yzlại có$:a^3+b^3=(a+b)(a^2-ab+b)^2\leq \sqrt{2(a^2+b^2)}[(a-b)^2+ab]=\sqrt{2[(z+y)^2+4yz]}(y+z)^2\leq \sqrt{4(y+z)^2}(y+z)^2=2(y+z)^
3(1)
tiếp:3(x+y)(y+z)(z+x)=12yz(y+z)\leq3(y+z)^2(y+z)=3(y+z)^3(2)$$(1)+(2)=đpcm$