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Question Number 202307 by MATHEMATICSAM last updated on 24/Dec/23

$Show that \frac{a^{3} + b^{3}}{a^{2} + b^{2}} > \frac{a^{2} + b^{2}}{a + b}$
	Answered by aleks041103 last updated on 24/Dec/23

	$(a^{3} + b^{3}) (a + b) - {(a^{2} + b^{2})}^{2} =$ $= a^{4} + a^{3} b + {a b}^{3} + b^{4} - a^{4} - b^{4} - 2 a^{2} b^{2} =$ $= a b (a^{2} + b^{2} - 2 a b) = a b {(a - b)}^{2}$ $\Rightarrow$ $a, b > 0 \land a \neq b \Rightarrow \frac{a^{3} + b^{3}}{a^{2} + b^{2}} > \frac{a^{2} + b^{2}}{a + b}$
	Answered by MM42 last updated on 24/Dec/23

	$i f a = - 1 & b = 2$ $\Rightarrow \frac{7}{5} > 5 i t s w r o n g .$
	Commented by aleks041103 last updated on 24/Dec/23

	$y e s .$ $f o r a a n d b d i f f e r e n t p o s i t i v e n u m b e r s$ $t h e i n e q u a l i t y i s t r u e$
	Answered by Frix last updated on 24/Dec/23

	$\frac{a^{3} + b^{3}}{a^{2} + b^{2}} - \frac{a^{2} + b^{2}}{a + b} > 0$ $\frac{a {(a - b)}^{2} b}{(a + b) (a^{2} + b^{2})} > 0$ $\frac{a b}{a + b} > 0$ $True if$ $a b > 0 \land a + b > 0 \lor a b < 0 \land a + b < 0$ $\Leftrightarrow$ $a > 0 \land b > 0 \lor a < 0 \land b < 0$
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