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

  • $\begingroup$ If you can't prove it, maybe it isn't true. Try looking for a counterexample. $\endgroup$ – Ragnar Dec 16 '13 at 12:52

Let $d=a-c$ and $e=a+b-c$ (so $b=e-d$)

Hence we have $$(2a-d+e)e<0$$

Consider $P(x)=ex^2+dx+a$

Hence $P(0)+P(-1)=2a-d+e$. So $P(0)+P(-1)$ is not of the same sign as $e$ ($e$ has the sign of the limit of $P$ when $|x|\rightarrow\infty$). So $P(x)$ has not always the same sign (positive or negative).

So P(X) has two distinct real roots.

Hence, $\Delta(P )=d^2-4ae$


  • $\begingroup$ why $P(x)$ has not always the same sign.I don't understand,can you post full ? Thank you $\endgroup$ – user94270 Dec 16 '13 at 13:27
  • $\begingroup$ Because $(2a-d+e)=P(0)+P(1)$ and $e=\lim_{x\rightarrow\infty}\frac{P(x)}{x^2}$ have not the same sign. $\endgroup$ – Xoff Dec 16 '13 at 13:28
  • $\begingroup$ you mean $\lim_{x\to \infty}P(x)\cdot[P(0)+P(1)]<0$,then why $P(x)$ has not sanme sign? $\endgroup$ – user94270 Dec 16 '13 at 13:31
  • $\begingroup$ yes, it means that at least $P(0)$ or $P(1)$ has not the same sign as the limit. $\endgroup$ – Xoff Dec 16 '13 at 13:32
  • $\begingroup$ and has not the same sign as $P(N)$ and $P(-N)$ for sufficient large $N$. $\endgroup$ – Xoff Dec 16 '13 at 13:37

$(2a+b)(a+b-c)<0 \implies$

$-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.


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