Prove-that-x-1-2-x-x-1-x-2-1-2-x-gt-0-

Question Number 158245 by HongKing last updated on 01/Nov/21

Prove that :

\frac{{(x - 1)}^{2}}{x} + \frac{x + 1}{\sqrt{x^{2} + 1}} ⩾ \sqrt{2}; \forall x > 0

Answered by mindispower last updated on 02/Nov/21

t \overset{f}{\to} 2 t g (t) i s b i j e c t i o n, [0, \frac{π}{2} [\overset{f}{\to} [0, + \infty [

x = t g (t) \Leftrightarrow

\frac{c o s (t)}{s i n (t)} (\frac{1 - 2 s i n (t) c o s (t)}{{c o s}^{2} (x)}) + s i n (t) + c o s (t) ⩾ \sqrt{2}

\frac{1}{c o s (t) s i n (t)} + s i n (t) + c o s (t) ⩾ 2 + \sqrt{2}, \forall t \in [0, \frac{π}{2} [

\frac{c o s (t)}{s i n (t)} + s i n (t) + \frac{s i n (t)}{c o s (t)} + c o s (t) ⩾ 2 + \sqrt{2}

a = s i n (t), b = c o s (t)

{\begin{cases} a^{2} + b^{2} = 1 \\ \frac{a}{b} + \frac{b}{a} + a + b \underset{. = f (a, b)}{∣} ⩾ 2 + \sqrt{2} \end{cases}

\Leftrightarrow \frac{1}{a b} + a + b ⩾ 2 + \sqrt{2}

\Leftrightarrow \frac{1}{a b} + a + b ⩾ 2 + 2

f (a, b) = \frac{1}{a b} + a + b - γ (a^{2} + b^{2} - 1)

\frac{1}{b} (- \frac{1}{a^{2}}) + 1 - 2 γ a = 0

\frac{1}{a} (- \frac{1}{b^{2}}) + 1 - 2 γ b = 0

\frac{1}{b} + \frac{1}{a} - 2 γ (\frac{a}{b} + \frac{b}{a}) = 0

b

γ = \frac{a + b}{2}

- \frac{1}{b^{2}} + a - b + \frac{1}{a^{2}} = 0

b^{2} - a^{2} + a^{2} b^{2} (a - b) = 0

- (a - b) (a + b + a^{2} b^{2}) = 0

\Rightarrow a = b = \frac{1}{\sqrt{2}}

M i n (f) = f (\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}) = \frac{1}{\frac{1}{2}} + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = 2 + \sqrt{2}

Commented by HongKing last updated on 02/Nov/21

perfect my dear Ser, thank you