Prove that $ab\leq \dfrac {1}{8}$ (Second Solution)

Given $b^2-4ac$ is a real root of equation :$ax^2+bx+c=0,\,\, (a\neq 0)$. Prove that $ab\leq \dfrac {1}{8}$.

Second solution is my solution: (I want to mention it here that the first solution is provided by a Taiwan friend of mine)

Multiply the equation of $ax^2+bx+c=0$ by $4a$, we get:

Prove that $ab\leq \dfrac {1}{8}$ (First Solution)

Given $b^2-4ac$ is a real root of equation :$ax^2+bx+c=0,\,\, (a\neq 0)$. Prove that $ab\leq \dfrac {1}{8}$.

Some students would find that this problem very confusing, as they have been told countless time that $b^2-4ac$  is actually a discriminant for the quadratic equation $ax^2+bx+c=0$.

In this instance, $b^2-4ac$ is both the discriminant and the root, they would think instantly something that looks like the following:

16x²y²-48x²y+24xy²+100x²+16y²-72xy+150x-48y+100=28

Determine the pair(s) of real numbers $(a,\,b)$ that satisfy the equation $16x^2y^2-48x^2y+24xy^2+100x^2+16y^2-72xy+150x-48y+100=28$.

At first glance,  this seems like we need to use the modular arithmetic method to solve for $(x,\,y)$ but wait a minute! We need to find not the integers but real for $(x,\,y)$, so nope, modular arithmetic is out of the question...

Find The Unknown Function...

Let $f(x)$ be an unknown function defined on $[0,\infty)$ with $f(0)=0$ and $f(x)\le x^2$ for all $x$. For each $0\le t$, let $A_t$ be the area of the region bounded by $y=x^2$, $y=ax^2$ (where $1<a$) and $y=t^2$. Let $B_t$ be the area of the region bounded by $y=x^2$, $y=f(x)$ and $x=t$. See the image below:

























a) Show that if $A_t=B_t$ for some time $t$, then:

[MATH]\int_0^{t^2}\sqrt{y}-\sqrt{\frac{y}{a}}\,dy=\int_0^t x^2-f(x)\,dx[/MATH]

b) Suppose $A_t=B_t$ for all $0\le t$. Find $f(x)$.

c) What is the largest value that $a$ can have so that $0\le f(x)$ for all $x$?

(2) Evaluate The Sum Of 1/xy+z-1+1/yz+x-1+1/xz+y-1

Regarding the problem to evaluate:

[MATH]\frac{1}{xy+z-1}+\frac{1}{yz+x-1}+\frac{1}{xz+y-1}[/MATH]

given:

[MATH]\begin{cases}x+y+z=2 \\[3pt] x^2+y^2+z^2=3 \\[3pt] xyz=4 \\ \end{cases}[/MATH]

you might be wondering if the proposed solution is the only way to stab at the problem, or more specifically, if you solved it differently, but in a more tedious or long way, can you be proud of it?