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Question Number 133443 by mnjuly1970 last updated on 22/Feb/21

$. . . . . . n i c e c a l c u l u s . . . . . . .$ $i f a, b, c ⩾ 0$ $a n d :: {a c o s}^{2} (x) + {b s i n}^{2} (x) ⩽ c$ $t h e n p r o v e t h a t ::$ $\sqrt{a} {c o s}^{2} (x) + \sqrt{b} {s i n}^{2} (x) ⩽ \sqrt{c}$ $. . . . . . . . . . . . .$
	Answered by mnjuly1970 last updated on 22/Feb/21

	${\vec{u}}_{1} = (\sqrt{a} c o s (x), \sqrt{b} s i n (x)) \in R^{2}$ ${\vec{u}}_{2} = (c o s (x), s i n (x)) \in R^{2}$ $∣ {\vec{u}}_{1} . {\vec{u}}_{2} ∣ \underset{s c h w a r t z}{\overset{c a u c h y}{⩽}} ∣ {\vec{u}}_{1} ∣∣ {\vec{u}}_{2} ∣$ $∣ \sqrt{a} {c o s}^{2} (x) + \sqrt{b} {s i n}^{2} (x) ∣⩽ \sqrt{{a c o s}^{2} (x) + {b s i n}^{2} (x)} . \sqrt{[{c o s}^{2} (x) + {s i n}^{2} (x)] = 1}$ $∣ \sqrt{a} {c o s}^{2} (x) + \sqrt{b} {s i n}^{2} (x) ∣⩽ \sqrt{c}$ $. . . . .$
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