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Question Number 145314 by loveineq last updated on 04/Jul/21

Let a, b ⩾ 0 and (a + 1) (b + 1) = {(a + b)}^{2} . Prove that

(a + b) \sqrt{{(a + 1)}^{3} + {(b + 1)}^{3}} ⩽ {(a + 1)}^{2} + {(b + 1)}^{2} ⩽ \frac{1}{2} [{(a + 1)}^{3} + {(b + 1)}^{3}]

Commented by justtry last updated on 05/Jul/21

(a + b) (b + 1) = {(a + b)}^{2}

\Leftrightarrow a b + a + b + 1 = a^{2} + 2 b + b^{2}

\Leftrightarrow 1 + (a + b) = a^{2} + 2 b + b^{2}

b = 0 \Rightarrow (1 + a) = a^{2}

a = 0 \Rightarrow (1 + b) = b^{2}

& {(1 + a)}^{3} ⩾ {(1 + a)}^{2} = {(a^{2})}^{2}

{(1 + b)}^{3} ⩾ {(1 + b)}^{2} = {(b^{2})}^{2}

\frac{1}{2} [{(1 + a)}^{3} + {(1 + b)}^{3}] ⩾ \frac{1}{2} [{(a^{2})}^{2} + {(b^{2})}^{2}] ⩾ \frac{{(a^{2} + b^{2})}^{2}}{4}

\Leftrightarrow \frac{1}{2} [{(1 + a)}^{3} + {(1 + b)}^{3}] ⩾ ({(a^{2})}^{2} + {(b^{2})}^{2}) ⩾ \frac{{(a^{2} + b^{2})}^{2}}{2}

\Leftrightarrow \frac{1}{2} [{(1 + a)}^{3} + {(1 + b)}^{3}] ⩾ \frac{1}{2} (a^{4} + b^{4} + 2 a^{2} b^{2}) = \frac{1}{2} [{(1 + a)}^{2} + {(1 + b)}^{2} + 2 {(a + b)}^{2}]

n o t e

\frac{1}{2} [{(1 + a)}^{2} + {(1 + b)}^{2} + 2 {(a + b)}^{2}] ⩾ {(1 + a)}^{2} + {(1 + b)}^{2} \to a = b = 1 o r a \neq b \neq 1, a, b ⩾ 0

s o

{\frac{1}{2} [{(1 + a)}^{3} + {(1 + b)}^{3}] ⩾ {(1 + a)}^{2} + (1 + b))}^{2}

\Leftrightarrow {(1 + a)}^{2} + {(1 + b)}^{2} ⩽ \frac{1}{2} [{(1 + a)}^{3} + {(1 + b)}^{3}]

\dots \dots \dots

Commented by loveineq last updated on 05/Jul/21

why 1 + a + b = a^{2} + 2 b + b^{2} ?