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Question Number 40306 by behi83417@gmail.com last updated on 19/Jul/18

Commented by MJS last updated on 20/Jul/18

$can we show the following ?$ $\frac{3}{2} ⩽ \frac{x^{2} + y z}{y^{2} + z^{2}} + \frac{y^{2} + x z}{x^{2} + z^{2}} + \frac{z^{2} + x y}{x^{2} + y^{2}} ⩽ \frac{21}{10}$
Commented by MJS last updated on 20/Jul/18
$f(x,y,z)=((x^2 +yz)/(y^2 +z^2 ))+((y^2 +xz)/(x^2 +z^2 ))+((z^2 +xy)/(x^2 +y^2 )) z=−(x+y) f(x,y,−(x+y))= =g(x,y)=((3(x^6 +y^6 +3xy(x^4 +y^4 )+8x^2 y^2 (x^2 +y^2 )+11x^3 y^3 ))/((x^2 +y^2 )((x+y)^2 +x^2 )((x+y)^2 +y^2 ))) (d/dx)[g(x,y)]=−((15xy^2 (2(x^8 −y^8 )+9xy(x^6 −y^6 )+16x^2 y^2 (x^4 −y^4 )+14x^3 y^3 (x^2 −y^2 )))/((x^2 +y^2 )^2 ((x+_ y)^2 +x^2 )^2 ((x+y)^2 +y^2 )^2 )) (d/dx)[g(x,y)]=0 ⇒ x=0 ∨ y=0 ∨ 2(x^8 −y^8 )+9xy(x^6 −y^6 )+16x^2 y^2 (x^4 −y^4 )+14x^3 y^3 (x^2 −y^2 )=0 x={−2y; −y; −(y/2); 0; y; −(y/2)(1+i(√3)); −(y/2)(1−i(√3))} f(x, y, z)={((21)/(10)); (3/2); ((21)/(10)); (3/2); ((21)/(10)); −6; −6} so these are the minima and maxima of f(x, y, z)$
$f (x, y, z) = \frac{x^{2} + y z}{y^{2} + z^{2}} + \frac{y^{2} + x z}{x^{2} + z^{2}} + \frac{z^{2} + x y}{x^{2} + y^{2}}$ $z = - (x + y)$ $f (x, y, - (x + y)) =$ $= g (x, y) = \frac{3 (x^{6} + y^{6} + 3 x y (x^{4} + y^{4}) + 8 x^{2} y^{2} (x^{2} + y^{2}) + 11 x^{3} y^{3})}{(x^{2} + y^{2}) ({(x + y)}^{2} + x^{2}) ({(x + y)}^{2} + y^{2})}$ $\frac{d}{d x} [g (x, y)] = - \frac{15 {x y}^{2} (2 (x^{8} - y^{8}) + 9 x y (x^{6} - y^{6}) + 16 x^{2} y^{2} (x^{4} - y^{4}) + 14 x^{3} y^{3} (x^{2} - y^{2}))}{{(x^{2} + y^{2})}^{2} {({(x +_{} y)}^{2} + x^{2})}^{2} {({(x + y)}^{2} + y^{2})}^{2}}$ $\frac{d}{d x} [g (x, y)] = 0 \Rightarrow x = 0 \lor y = 0 \lor$ $2 (x^{8} - y^{8}) + 9 x y (x^{6} - y^{6}) + 16 x^{2} y^{2} (x^{4} - y^{4}) + 14 x^{3} y^{3} (x^{2} - y^{2}) = 0$ $x = {- 2 y; - y; - \frac{y}{2}; 0; y; - \frac{y}{2} (1 + i \sqrt{3}); - \frac{y}{2} (1 - i \sqrt{3})}$ $f (x, y, z) = {\frac{21}{10}; \frac{3}{2}; \frac{21}{10}; \frac{3}{2}; \frac{21}{10}; - 6; - 6}$ $so these are the minima and maxima of f (x, y, z)$
	Answered by tanmay.chaudhury50@gmail.com last updated on 19/Jul/18

	$k = \frac{x y}{x^{2} + y^{2}} + \frac{y z}{y^{2} + z^{2}} + \frac{x z}{x^{2} + z^{2}} + \frac{z^{2}}{x^{2} + y^{2}} + \frac{x^{2}}{y^{2} + z^{2}} + \frac{y^{2}}{x^{2} + z^{2}}$ $\frac{x^{2} + y^{2}}{2} ⩾ \sqrt{x^{2} y^{2}}$ $\frac{x^{2} + y^{2}}{x y} ⩾ 2 s o \frac{x y}{x^{2} + y^{2}} ⩽ \frac{1}{2}$ $\frac{z^{2}}{x^{2} + y^{2}} z = - (x + y) s o z^{2} = x^{2} + y^{2} + 2 x y$ $\frac{z^{2}}{x^{2} + y^{2}} = 1 + \frac{2 x y}{x^{2} + y^{2}}$ $1 + \frac{2 x y}{x^{2} + y^{2}} ⩽ 2$ $k ⩽ \frac{1}{2} + \frac{1}{2} + \frac{1}{2} + 2 + 2 + 2$ $k ⩽ \frac{15}{2}$
	Answered by math1967 last updated on 19/Jul/18

	$x, y, z \in R ∴ {(x - y)}^{2} ⩾ 0$ ${(y - z)}^{2} ⩾ 0, {(z - x)}^{2} ⩾ 0$ $∴ x^{2} + y^{2} ⩾ 2 x y . . . . . . . 1$ $F r o m 1 \frac{1}{x^{2} + y^{2}} ⩽ \frac{1}{2 x y}$ $\frac{x y}{x^{2} + y^{2}} ⩽ \frac{1}{2} . . . . . . (2)$ $A g a i n x + y + z = 0$ $\Rightarrow x^{2} + y^{2} + 2 x y = z^{2}$ $∴ x^{2} + y^{2} + 3 x y = x y + z^{2}$ $F r o m (2) \frac{3 x y}{x^{2} + y^{2}} ⩽ \frac{3}{2}$ $1 + \frac{3 x y}{x^{2} + y^{2}} ⩽ \frac{5}{2}$ $\frac{x^{2} + y^{2} + 3 x y}{x^{2} + y^{2}} ⩽ \frac{5}{2}$ $\frac{x y + z^{2}}{x^{2} + y^{2}} ⩽ \frac{5}{2} [x^{2} + y^{2} + 3 x y = x y + z^{2}]$ $s i m i l a r l y \frac{y z + x^{2}}{y^{2} + z^{2}} ⩽ \frac{5}{2}$ $\frac{z x + y^{2}}{z^{2} + x^{2}} ⩽ \frac{5}{2}$ $∴ \frac{x y + z^{2}}{x^{2} + y^{2}} + \frac{y z + x^{2}}{y^{2} + z^{2}} + \frac{z x + y^{2}}{z^{2} + x^{2}} ⩽ \frac{15}{2}$
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