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x-y-z-R-a-b-c-gt-0-prove-that-x-2-a-y-2-b-z-2-c-x-y-z-2-a-b-c-




Question Number 201615 by hardmath last updated on 09/Dec/23
x,y,z ∈ R  a,b,c>0  prove that:  (x^2 /a) + (y^2 /b) + (z^2 /c) ≥ (((x + y + z)^2 )/(a + b + c))
$$\mathrm{x},\mathrm{y},\mathrm{z}\:\in\:\mathbb{R} \\ $$$$\mathrm{a},\mathrm{b},\mathrm{c}>\mathrm{0} \\ $$$$\mathrm{prove}\:\mathrm{that}: \\ $$$$\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{a}}\:+\:\frac{\mathrm{y}^{\mathrm{2}} }{\mathrm{b}}\:+\:\frac{\mathrm{z}^{\mathrm{2}} }{\mathrm{c}}\:\geqslant\:\frac{\left(\mathrm{x}\:+\:\mathrm{y}\:+\:\mathrm{z}\right)^{\mathrm{2}} }{\mathrm{a}\:+\:\mathrm{b}\:+\:\mathrm{c}} \\ $$
Answered by AST last updated on 09/Dec/23
(√(((x/( (√a))))^2 +((y/( (√b))))^2 +((z/( (√c))))^2 ))(√(((√a))^2 +((√b))^2 +((√c))^2 ))  ≥(x+y+z)  ∣u∣∣v∣≥u∙v where u=⟨(x/( (√a))),(y/( (√b))),(z/( (√c)))⟩;v=⟨(√a),(√b),(√c)⟩
$$\sqrt{\left(\frac{{x}}{\:\sqrt{{a}}}\right)^{\mathrm{2}} +\left(\frac{{y}}{\:\sqrt{{b}}}\right)^{\mathrm{2}} +\left(\frac{{z}}{\:\sqrt{{c}}}\right)^{\mathrm{2}} }\sqrt{\left(\sqrt{{a}}\right)^{\mathrm{2}} +\left(\sqrt{{b}}\right)^{\mathrm{2}} +\left(\sqrt{{c}}\right)^{\mathrm{2}} } \\ $$$$\geqslant\left({x}+{y}+{z}\right) \\ $$$$\mid{u}\mid\mid{v}\mid\geqslant{u}\centerdot{v}\:{where}\:{u}=\langle\frac{{x}}{\:\sqrt{{a}}},\frac{{y}}{\:\sqrt{{b}}},\frac{{z}}{\:\sqrt{{c}}}\rangle;{v}=\langle\sqrt{{a}},\sqrt{{b}},\sqrt{{c}}\rangle \\ $$

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