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Let $0 < x < 1 .$ Prove that $$\frac{2x^2}{1 - x}+\frac{(1 - x)^2}{x} \ge \sqrt{9+6\sqrt{3}}-3$$

My work

$$\frac{2x^2}{1 - x}+\frac{(1 - x)^2}{x}$$

$$\frac{2x^2}{1 - x} + \frac{(1 - x)^2}{x} = \frac{2x^3}{x(1 - x)} + \frac{(1 - x)^3}{x(1 - x)}$$

$$\frac{2x^3 + (1 - x)^3}{x(1 - x)}$$

Now I expand the cube

$$\frac{2x^3 + (1 - x)^3}{x(1 - x)}$$

$$\frac{2x^3 + 1 - 3x + 3x^2 - x^3}{x(1 - x)}$$

$$\frac{(x^3 - 3x^2 - 3x + 1)}{x(1 - x)}$$

Am I on right way?

Martin.s
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  • Oh it's a typo(let me edit) and what to do next ? – Martin.s Sep 14 '23 at 17:27
  • For this kind of problem, one might try to compute the derivative of the function defined by $x\mapsto \frac{2x^2}{1-x}+\frac{(1-x)^2}{x}$, find the critical points on $]0,1[$ and sign of the derivative, hence getting the variations of the original function, then you'd be able to see where the function attains a minimum, and what that minimum is. Although in this particular case, I don't know how hard/easy the computations would be as I have not tried it. – t_kln Sep 14 '23 at 18:13
  • If you have done calculus you can differentiate and set numerator = 0 and solve. – mathreadler Sep 14 '23 at 18:13
  • Yeah a very big thanks I got your point – Martin.s Sep 14 '23 at 18:14

2 Answers2

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$\text{LHS} = -2-2x+\frac2{1-x}+\frac1x-2+x = \frac1x-x+\frac2{1-x}-4$

For $p,q>0$ write $\text{LHS}=\frac1x+px+q(1-x)+\frac2{1-x}-4-q$

Solution by AM-GM provided $px^2=1, q(1-x)^2=2, p-q=-1 \implies \frac1{x^2}-\frac2{(1-x)^2}=-1$

$x^4 - 2 x^3 - 2 x + 1=0 \iff (x^2-x+1)^2=3x^2 \implies x=\frac{1-\sqrt2\sqrt[4]3+\sqrt3}2\in(0,1)$

$\text{LHS} \ge \frac1x-x+\frac2{1-x}-4 =\sqrt{9 + 6\sqrt3}-3$

Martin.s
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If you want to use calculus you can differentiate the function $$x\to f(x) = \frac{2x^2}{1 - x}+\frac{(1 - x)^2}{x}$$

We get $$f'(x)=\frac{-x^4+2x^3+2x-1}{x^2(1-x)^2}$$

Now if we solve $f'(x) = 0$ on $x\in[0,1]$ we will find any extremum there.

This is as easy as solving the quartic equation $-x^4+2x^3+2x-1 = 0$ because the denominator can not affect nullness of the expression unless it is 0 but as we can se from it's factorization this only occurs at the end points of the interval.

A quartic equation can be solved algebraically or numerically. I will leave the algebraic solution as an exercise.

A few Newton iterations will for example bring us the approximation for the position of minimum $x = 0.4354205446823390$. We can calculate the function value at this point $1.403669475041612$ which up to machine precision is equal to the prescribed value $\sqrt{9+6\cdot\sqrt{3}}-3$

A dampened fix point iteration also works in finding the same approximation:

$x_{n+1} = 0.5x_n + 0.5g(x_n), g(x) = ((x^4-2x+1)/2)^{1/3}$

mathreadler
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