Inequality with x,y

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Inequality with x,y

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#1 h Apr 20, 2025, 11:46 AM

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Let $x,y\ge 0$ such that $5(x^3+y^3) \leq 16(1+xy)$ . Prove that:
$8+xy\geq 3(x+y)$

This post has been edited 1 time. Last edited by GeoMorocco, Apr 20, 2025, 11:46 AM

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#3 h Apr 28, 2025, 10:29 AM

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$5(x^3+y^3) \le 16(1+xy)$ ;

or $5(x+y)(x^2-xy+y^2) \le 16(1+xy)$ ;

or $3(x+y) \le \frac{48}{5} \cdot \frac{8+xy-7}{x^2-xy+y^2}$ ;

or $3(x+y) \le \frac{48(8+xy) - 336}{5(x^2-xy+y^2)}$ ;

or $3(x+y) \le \left[\frac{48}{5(x^2-xy+y^2)}+1-1\right](8+xy) - \frac{336}{5(x^2-xy+y^2)}$ ;

or $3(x+y) \le 8+xy + \frac{48(8+xy)-5(x^2-xy+y^2)(8+xy)-336}{5(x^2-xy+y^2)}$ ;

or $3(x+y) \le 8+xy - \frac{336 +[5((x-y)^2+xy)-48](8+xy)}{5[(x-y)^2+xy]} \le 8+xy$ ;

or $3(x+y) \le 8+xy$ for all real $x,y \ge 0$ .

QED

This post has been edited 2 times. Last edited by Mathzeus1024, Apr 28, 2025, 10:34 AM

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