Algebraic Relationships - Quadratic Equations

Question

I am having a tough time with the following question:

If $x$ is real and $p=3(x^2 + 1)/(2x-1)$, then prove that $p^2 - 3(p+1)\geq 0$.

I don't know how to tackle this question.

Thanks for your help! :)

Answer 1

The given expression can be re-written as $$3x^2-2px+(3+p)=0.$$ For this to have real solutions, we need the discriminant to be non-negative, i.e. $$4p^2-12(3+p) \geq 0.$$ This is same as $p^2-3(p+3) \geq 0.$

From this your inequality follows as well.

Answer 2

The "hard" way to tackle it: evaluate the given condition from the definition of $p$. $$p^2 - 3(p+1)= (3\frac{x^2 + 1}{2x-1})^2+3(3\frac{x^2 + 1}{2x-1}+1)=\frac{9x^4-18x^3+15x^2-6x+15}{(2x-1)^2}\ge0.$$

By plotting the polynomial at the numerator, you see that it has no real root. Unfortunately, you need to prove this. Find the minimum by cancelling the first derivative:$$36x^3-54x^2+30x-6=0.$$Plotting again, you see a single root at $x=\frac12$. The value at the minimum is positive, $ \frac{225}{16}$.

For complete rigor, you have to prove that the derivative has no other root. This is done from the Cardano's formulas.

Answer 3

If $$ p=3\frac{x^2+1}{2x-1} $$ then $$ \begin{align} &p^2-3p-3\\[6pt] &=9\frac{x^4+2x^2+1}{4x^2-4x+1}-9\frac{2x^3-x^2+2x-1}{4x^2-4x+1}-3\frac{4x^2-4x+1}{4x^2-4x+1}\\ &=3\frac{3x^4-6x^3+5x^2-2x+5}{4x^2-4x+1}\\ &=\frac3{(2x-1)^2}\big[3(x^2-x)^2+(x-1)^2+x^2+4\big]\\[3pt] &\ge0 \end{align} $$

---

Further Analysis

I noticed that $x^2-x$ and $(x-1)^2+x^2$ both have a minimum at $x=\frac12$, therefore, we can write everything in terms of $x-\frac12$: $$ \begin{align} &\color{#C00000}{3(x^2-x)^2}+\color{#00A000}{(x-1)^2+x^2+4}\\ &=\color{#C00000}{3((x-\tfrac12)^2-\tfrac14)^2}+\color{#00A000}{2(x-\tfrac12)^2+\tfrac92}\\ &=\color{#C00000}{3(x-\tfrac12)^4-\tfrac32(x-\tfrac12)^2+\tfrac3{16}}+\color{#00A000}{2(x-\tfrac12)^2+\tfrac92}\\ &=3(x-\tfrac12)^4+\tfrac12(x-\tfrac12)^2+\tfrac{75}{16}\\ &=\tfrac3{16}(2x-1)^4+\tfrac2{16}(2x-1)^2+\tfrac{75}{16} \end{align} $$ Therefore, using the results from above, $$ p^2-3p-3=\frac3{16}\left[3(2x-1)^2+2+\frac{75}{(2x-1)^2}\right]\\ $$ Since $3(2x-1)^2\cdot\dfrac{75}{(2x-1)^2}=225$, we have that $$ 3(2x-1)^2+\dfrac{75}{(2x-1)^2}\ge2\sqrt{225}=30 $$ with equality if and only if $3(2x-1)^2=\dfrac{75}{(2x-1)^2}$; that is, $x=\dfrac{1\pm\sqrt5}2$.

Therefore, $$ p^2-3p-3\ge6 $$ with equality if and only if $x=\dfrac{1\pm\sqrt5}2$.