A geometric inequality in a triangle

Question

If $x,y,z>0,$ could it be:

$ \frac { z(2z+x)}{z+x } b^2+\frac { y(2y+x) }{y+x } c^2> \frac {yz}{y+z} a^2$

for a triangle $ABC,$ with $a=BC,$ $b=AC,$ $c=AB?$

If $b^2+c^2=a^2$, then the inequality holds, since

$ \frac { z(2z+x)}{z+x } - \frac {yz}{y+z} =\frac{ z^2 (x+y+2z)}{ (z+x)(y+z)}, \frac { y(2y+x)}{y+x } - \frac {yz}{y+z} =\frac{ y^2 (x+2y+z)}{ (z+x)(y+z)} $.

If $b^2+c^2>a^2$, then the inequality holds, since

$ \frac { z(2z+x)}{z+x } b^2+\frac { y(2y+x) }{y+x } c^2> \frac{yz}{y+z} b^2 + \frac{yz}{y+z} c^2 > \frac {yz}{y+z} a^2$

I can’t handle the case $b^2+c^2<a^2$ and I need some help

Answer 1

By C-S $$ \frac { z(2z+x)}{z+x } b^2+\frac { y(2y+x) }{y+x } c^2\geq\frac{(b+c)^2}{\frac{z+x}{z(2z+x)}+\frac{y+x}{y(2y+x)}}>$$ $$>\frac{a^2}{\frac{z+x}{z(2z+x)}+\frac{y+x}{y(2y+x)}}>\frac{a^2}{\frac{z+x}{z(z+x)}+\frac{y+x}{y(y+x)}}=\frac{a^2yz}{y+z}.$$