Question-165795

Question Number 165795 by mnjuly1970 last updated on 08/Feb/22

Answered by ArielVyny last updated on 08/Feb/22

s o i t f (x, y) = x^{2} + \frac{1}{x^{2}} + y^{2} + \frac{y}{x}

m o n t r o n s q u e f (x, y) ⩾ \sqrt{3} x \neq 0

- f i x o n s x \in R o n a f_{x} (y) = x^{2} + \frac{1}{x^{2}} + y^{2} + y \frac{1}{x}

f_{x} (y) = {(y + \frac{1}{2 x})}^{2} - \frac{1}{4 x^{2}} + x^{2} + \frac{1}{x^{2}}

= {(y + \frac{1}{2 x})}^{2} + \frac{3}{4 x^{2}} + x^{2}

α = \frac{1}{2 x} \to x = \frac{1}{2 α}

f_{x} (y) = {(y + α)}^{2} + 3 α^{2} + \frac{1}{4 α^{2}}

p o s o n s α^{2} \in [0; 1] o n a 3 α^{2} + \frac{1}{4 α^{2}} ⩾ \sqrt{3}

d e m e m e p o u r α^{2} \in] 1; + \infty [3 α^{2} + \frac{1}{4 α^{2}} ⩾ \sqrt{3}

d e p l u s c b a q u e t e r m e d e f_{x} (y) e s t p o s i t i f d o n c

f_{x} (y) ⩾ \sqrt{3}

- f i x o n s y f_{y} (x) = x^{2} + \frac{1}{x^{2}} + \frac{y}{x} + y^{2} (x \neq 0)

f_{y} (x) = {(x - \frac{1}{x})}^{2} + y^{2} + 2 + \frac{y}{x}

= {(x - \frac{1}{x})}^{2} + {(y + \frac{1}{2 x})}^{2} - \frac{1}{2 x^{2}} + 2

p a r r a i s o n n e m e n t a n a l o g u e o n t r o u v e l^{'} i n e g a l i t e

N B o n p o u r r a e t u d i e r q (x) = - \frac{1}{2 x^{2}} + 2 - \sqrt{3}

e x t r a i r e l^{'} i n e r v a l o u q (x) e s t n e g a t i f

e t m o n t r e r q u e s i l^{'} o n a j o u t e {(x - \frac{1}{x})}^{2} + {(y + \frac{1}{2 x})}^{2}

f_{y} (x) ⩾ \sqrt{3}

Commented by mnjuly1970 last updated on 09/Feb/22

v e r y n i c e t h a n k y o u s i r

Answered by mr W last updated on 08/Feb/22

x^{2} + \frac{1}{x^{2}} + y^{2} + \frac{y}{x}

= x^{2} + \frac{3}{4 x^{2}} + \frac{1}{4 x^{2}} + \frac{y}{x} + y^{2}

= x^{2} + \frac{3}{4 x^{2}} + {(\frac{1}{2 x} + y)}^{2}

⩾ x^{2} + \frac{3}{4 x^{2}}

⩾ 2 \sqrt{x^{2} \times \frac{3}{4 x^{2}}} = \sqrt{3}

Commented by mnjuly1970 last updated on 09/Feb/22

t h a n k s a l o t s i r W