<s>[INST] How prove this $(a-c)^2-4a(a+b-c)>0$ Question:
let $a,b,c$ are real numbers,and such
$$(2a+b)(a+b-c)<0$$
prove or disprove 
$$(a-c)^2>4a(a+b-c)$$
My try: since
$$(a-c)^2-4a(a+b-c)=a^2-2ac+c^2-4a^2-4ab+4ac=a^2+2ac+c^2-4ab-4a^2=(a+c)^2-4a(a+b)$$
Then I can't,Thank you
 [/INST] $(2a+b)(a+b-c)<0 \implies$ 
$-2a<b<c-a $ in case $c>-a$ <1> 
or $c-a<b<-2a$ in case $c<-a$ <2>
case <1> 
if $a>0,b+a<c,-4a(a+b)>-4ac$, 
if $a<0, b+a>-2a,-4a(a+b)>8a^2$
case<2> is similar.
</s>