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Question Number 204621 by hardmath last updated on 23/Feb/24

a, b, c \in R^{+}

If \sqrt{a} + \sqrt{b} + \sqrt{c} = 1

Prove that : a + b + c ⩾ \frac{1}{3}

Answered by A5T last updated on 23/Feb/24

\frac{a + b + c}{3} ⩾ {(\frac{\sqrt{a} + \sqrt{b} + \sqrt{c}}{3})}^{2} = \frac{1}{9} \Rightarrow a + b + c ⩾ \frac{1}{3}

Answered by mr W last updated on 23/Feb/24

a + b + c = \frac{{(\sqrt{a})}^{2}}{1} + \frac{{(\sqrt{b})}^{2}}{1} + \frac{{(\sqrt{c})}^{2}}{1}

⩾ \frac{{(\sqrt{a} + \sqrt{b} + \sqrt{c})}^{2}}{1 + 1 + 1} = \frac{1^{2}}{3} = \frac{1}{3}

Commented by mr W last updated on 23/Feb/24

Answered by mr W last updated on 24/Feb/24

Commented by mr W last updated on 24/Feb/24

s a y \sqrt{a} = x, \sqrt{b} = y, \sqrt{c} = z

x + y + z = 1 i s t h e p l a n e Δ A B C .

P (x, y, z) i s a p o i n t o n t h e p l a n e .

a + b + c = x^{2} + y^{2} + z^{2} = {O P}^{2}

{O P}_{m i n} i s t h e d i s t a n c e d f r o m O t o t h e

p l a n e Δ A B C .

{O P}_{m i n} = d = \frac{∣ 0 + 0 + 0 - 1 ∣}{\sqrt{1^{2} + 1^{2} + 1^{2}}} = \frac{1}{\sqrt{3}}

{O P}_{m i n}^{2} = \frac{1}{3}

\Rightarrow a + b + c = {O P}^{2} ⩾ {O P}_{m i n}^{2} = \frac{1}{3}

Commented by hardmath last updated on 24/Feb/24

Perfect solution, thank you dear professor