Prove the following inequality: $a^2+c^2-b^2>\frac{a^2c^2}{b^2}$

Question

Suppose we the following real numbers:

{$a$,$b$,$c$} $\in \mathbb{R}^+$, where $a>b>c$. Is there any way to prove the following:

$a^2+c^2-b^2>\frac{a^2c^2}{b^2}$

Answer 1

Let $a=1,\;c=-1$.

Then as $b \to 0^{+}$, the LHS approaches $2$, while the RHS approaches $+\infty$.

Edit:

My answer above was based on the original question, with $a,b,c$ unrestricted in sign.

If we assume $a > b > c > 0$, we can argue as follows . . .

Let $x = a/b,\; y = c/b$. \begin{align*} \text{Then}\;\;&a^2+c^2-b^2>\frac{a^2c^2}{b^2}\\[4pt] \iff\;&\frac{a^2+c^2-b^2}{b^2}>\frac{a^2c^2}{b^4}\\[4pt] \iff\;&x^2 + y^2 - 1>x^2y^2\\[6pt] \iff\;&x^2y^2-x^2-y^2 + 1 < 0\\[6pt] \iff\;&(x^2-1)(y^2-1) < 0\\[4pt] \end{align*} which is true since $x > 1$ and $0 < y < 1$.

Answer 2

This inequality is equivalent to $$a^2b^2+b^2c^2-b^4-a^2c^2>0.$$ Can you prove or refute this?

Answer 3

Let $b=c+u$ and $a=c+u+v$, where $u$ and $v$ are positives.

Hence, $$b^2(a^2+c^2-b^2)-a^2c^2=4uvc^2+2uv(3u+v)c+u^2v(2u+v)>0.$$

PS. Now I see that $$b^2(a^2+c^2-b^2)-a^2c^2=-b^4+a^2b^2+c^2b^2-a^2c^2=(a^2-b^2)(b^2-c^2)>0.$$

Answer 4

The special case. Let $a$, $b$, and $c$ be the sides of a triangle such that angle $(bc)=90^o$. Then,

$$a^2 + c^2 - b^2 - 2ac \cos(ac) > \frac{a^2c^2}{b^2} - 2ac\cos(ac),$$ $$0 > \frac{a^2c^2}{b^2} - 2ac\frac{c}{a} = \frac{a^2c^2}{b^2} - 2c^2.$$

As $b \to a$, the RHS of the inequality is $c^2 - 2 c^2 <0$. As $b \to c$, the RHS of the inequality is $a^2 - 2c^2 =0$ (this is because $\cos(45^o) = 1/{\sqrt2}$). Due to the continuity of the RHS w.r.t $b$, the RHS must be strictly negative for values of $a>b>c$.

May be you can generalize it from here.