If-a-2-b-2-2-Then-2a-7-2b-7-6-

Question Number 203149 by hardmath last updated on 11/Jan/24

If a^{2} + b^{2} = 2

Then \sqrt{2 a + 7} + \sqrt{2 b + 7} ⩽ 6

Commented by mr W last updated on 12/Jan/24

\underset{a = b = - 1}{2 \sqrt{5}} ⩽ \sqrt{2 a + 7} + \sqrt{2 b + 7} ⩽ \underset{a = b = 1}{6}

Answered by AST last updated on 11/Jan/24

L e t p = 2 a + 7; q = 2 b + 7

\Rightarrow p^{2} = 4 a^{2} + 28 a + 49; q^{2} = 4 b^{2} + 28 b + 49

\Rightarrow p^{2} + q^{2} = 4 (a^{2} + b^{2}) + 28 (a + b) + 98 = 106 + 28 (a + b)

⩽ 106 + 28 (2 \sqrt{\frac{a^{2} + b^{2}}{2}}) = 162

\Rightarrow 81 ⩾ \frac{p^{2} + q^{2}}{2} ⩾ {(\frac{p + q}{2})}^{2} \Rightarrow 9 ⩾ \sqrt{\frac{p^{2} + q^{2}}{2}} ⩾ \frac{p + q}{2} ⩾ {(\frac{\sqrt{p} + \sqrt{q}}{2})}^{2}

\Rightarrow {(\frac{\sqrt{p} + \sqrt{q}}{2})}^{2} ⩽ 9 \Rightarrow \sqrt{p} + \sqrt{q} = \sqrt{2 a + 7} + \sqrt{2 b + 7} ⩽ 6

Commented by hardmath last updated on 12/Jan/24

thank you dear professors

Answered by witcher3 last updated on 11/Jan/24

f (x) = x^{\frac{1}{2}} is concave f^{″} (x) = - \frac{1}{4} x^{- \frac{3}{2}} < 0

\Rightarrow f (\frac{a + b}{2}) ⩾ \frac{1}{2} f (a) + \frac{1}{2} f (b)

\Rightarrow \sqrt{2 a + 7} + \sqrt{2 b + 7} ⩽ 2 \sqrt{\frac{2 a + 7 + 2 b + 7}{2}}

⩽ 2 \sqrt{a + b + 7}

{(a + b)}^{2} ⩽ 2 (a^{2} + b^{2}) \Rightarrow a + b ⩽ \sqrt{2 (a^{2} + b^{2})} = 2

\Rightarrow LhS ⩽ 2 . \sqrt{2 + 7} = 2.3 = 6

Facebook Tweet Pin