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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

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