Question-160375

Question Number 160375 by HongKing last updated on 28/Nov/21

Commented by ghimisi last updated on 28/Nov/21

\frac{1}{2 b + 1} o r \frac{1}{4 b + 1}

Commented by HongKing last updated on 28/Nov/21

Sorry my dear Ser, \frac{1}{4 b + 1}

Answered by ghimisi last updated on 28/Nov/21

\forall t \in R, (t^{2} - t + 1) (t^{2} - 2 t + 1) ⩾ 0 \Rightarrow

t^{4} + 4 t^{2} + 1 ⩾ 3 t^{3} + 3 t

t = \frac{x^{a}}{x^{b}} \Rightarrow x^{4 a} + 4 x^{2 a + 2 b} + x^{4 b} ⩾ 3 x^{3 a + b} + 3 x^{a + 3 b}

\Rightarrow \underset{0}{\int^{1}} x^{4 a} d x + 4 \underset{0}{\int^{1}} x^{2 a + 2 b} d x + \underset{0}{\int^{1}} x^{4 b} d x ⩾ 3 \underset{0}{\int^{1}} x^{3 a + b} d x + 3 \underset{0}{\int^{1}} x^{a + 3 b} d x

\Rightarrow \frac{1}{4 a + 1} + \frac{4}{2 a + 2 b + 1} + \frac{1}{4 b + 1} ⩾ \frac{3}{3 a + b + 1} + \frac{3}{a + 3 b + 1}

Commented by HongKing last updated on 28/Nov/21

Cool thank you so much my dear Ser