inequalities - 5/4

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inequalities - 5/4

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pennypc123456789

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pennypc123456789

#1 h Apr 5, 2025, 8:57 AM

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Given real numbers $x, y$ satisfying $|x| \le 3, |y| \le 3$ . Prove that:
$0 \le (x^2 + 1)(y^2 + 1) + 4(x - 1)(y - 1) \le 164.$

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#2 h Apr 5, 2025, 10:40 AM

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Let us expand to obtain $f(x,y) = x^{2}y^{2}+x^2+y^2+4xy-4x-4y+5$ for all real $x,y \in [-3,3]$ . Taking $\nabla f= 0$ yields:

$f_{x} = 2xy^2+2x+4y-4=0$ (i);

$f_{y} = 2x^{2}y + 2y +4x-4=0$ (ii)

of which subtracting (ii) from (i) produces:

$2xy(y-x)-2(y-x)+4(y-x) = 0 \Rightarrow (y-x)(2+2xy) = 0 \Rightarrow y=x$ (iii) or $y=-\frac{1}{x}$ (iv). Substitution of (iii) into either of (i) or (ii) yields:

$2x^3+6x-4=0 \Rightarrow x = 0.59607; -0.2980 \pm 1.8073i \Rightarrow (x,y) = (0.59607, 0.59607)$ (v);

and the same with (iv) gives:

$2x-4-\frac{2}{x}=0 \Rightarrow 2x^2-4x-2=0 \Rightarrow x = \frac{4 \pm \sqrt{16-4(2)(-2)}}{2(2)} \Rightarrow (x,y) = (1 \pm \sqrt{2}, 1\mp \sqrt{2})$ (vi).

Checking the Hessian matrix $F(x,y) = \begin{bmatrix} f_{xx} && f_{xy} \\ f_{yx} && f_{yy} \end{bmatrix} = \begin{bmatrix} 2y^2+2 && 4xy+4 \\ 4xy+4 && 2x^2+2 \end{bmatrix}$ yields:

$|F(0.59607, 0.59607)| = \begin{vmatrix} 2(0.59607)^2+2 && 4(0.59607)^2+4 \\ 4(0.59607)^2+4 && 2(0.59607)^2+2 \end{vmatrix} = -22.042 < 0 \Rightarrow$ a maximum.

$|F(1 \pm \sqrt{2}, 1 \mp \sqrt{2})| = \begin{vmatrix} 2(1 \mp \sqrt{2})^2+2 && 4(1 \pm \sqrt{2})(1\mp \sqrt{2})+4 \\ 4(1 \pm \sqrt{2})(1\mp \sqrt{2})+4 && 2(1 \pm \sqrt{2})^2+2 \end{vmatrix} = 32 > 0 \Rightarrow$ a minimum.

Hence:

$\textcolor{red}{f_{MIN} = f(1 \pm \sqrt{2}, 1 \mp \sqrt{2}) = 0}$ and $\textcolor{red}{f_{MAX} = f(0.59607, 0.59607) = 2.49}$ ;

which proves that $0 \le f_{MIN} < f_{MAX} \le 164$ .

This post has been edited 3 times. Last edited by Mathzeus1024, Apr 5, 2025, 10:42 AM

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sqing

#3 h Apr 5, 2025, 11:35 AM • 1 Y

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Canada Repêchage 2017:
Prove for all real numbers $x, y$ , $(x^2 + 1)(y^2 + 1) + 4(x - 1)(y - 1) \geq 0.$ Determine when equality holds.

This post has been edited 1 time. Last edited by sqing, Apr 5, 2025, 11:36 AM

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