Let $ x,y,z>0 $ and $xy+z^2=8$. Prove that $$\frac{x+y}{z}+\frac{y+z}{x^2}+\frac{z+x}{y^2}\geq 4$$

lets take k=√xy. by AmGm (x+y)/z>=2k/z and (y+z)/x^2+(z+x)/y^2=(z(x^2+y^2)+x^3+y^3)/(xy)^2>=(2zk^2+2k^3)/k^4=2(z+k)/k^2 and now we only have to find smallest possible value of (2k/z)+2(z+k)/k^2 where k^2+z^2=8. For first we will prove that k+z>=kz. (k+z)^2=k^2+z^2+2kz=8+2kz>=(kz)^2 8+2kz>=(kz)^2 8>=kz(kz-2) by AmGm 8=k^2+z^2>=2kz 8>=2kz 4>=kz 8=4(4-2)>=kz(kz-2) now we back to the finding the smallest value of 2((k/z)+(z+k)/k^2) and again by AmGm 2((k/z)+(z+k)/k^2)>=2*2√((k(z+k)/zk^2)=2*2√((z+k)/zk)>=2*2=4 (the statment √((z+k)/zk)>=1 comes from z+k>=zk that we already have proved) so the answear is 4 and the example is x=y=z=2

sqing wrote: Let $ x,y,z>0 $ and $xy+z^2=8$. Prove that $$\frac{x+y}{z}+\frac{y+z}{x^2}+\frac{z+x}{y^2}\geq 4$$