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Question Number 143516 by mathdanisur last updated on 15/Jun/21

Commented by amin96 last updated on 15/Jun/21

$?$
	Answered by MJS_new last updated on 15/Jun/21

	$trivial solutions$ $x = y = z = 0$ $x = y = z = 1$
	Commented by mathdanisur last updated on 15/Jun/21

	$t h a n k s S i r, i s a c o m p r e h e n s i v e$ $s o l u t i o n p o s s i b l e, p l e a s e . .$
	Answered by mindispower last updated on 15/Jun/21
	$y≥0,z≥0,x≥0.We applie AM−GM 1⇒ 1+x^2 +x^4 +x^6 ≥4x^3 ⇒4x^4 ≥4x^3 y⇒y≤x 2⇒ 1+y^2 +y^4 +y^6 +y^8 ≥5y^4 ⇒5y^5 ≥5y^4 z ⇒z≤y 3⇒z≥x⇒(1,2,3)⇒x=y=z x≠0 1⇔4x^3 =x^6 +x^4 +x^2 +1.AM−GM equalitie ⇒1=x=x^2 =x^4 =x^6 ⇒(x,y,z)=(1,1,1) x=0⇒y=0,z=0 (x,y,z)∈{(1,1,1);(0,0,0)}$
	$y ⩾ 0, z ⩾ 0, x ⩾ 0 . W e a p p l i e A M - G M$ $1 \Rightarrow$ $1 + x^{2} + x^{4} + x^{6} ⩾ 4 x^{3} \Rightarrow 4 x^{4} ⩾ 4 x^{3} y \Rightarrow y ⩽ x$ $2 \Rightarrow$ $1 + y^{2} + y^{4} + y^{6} + y^{8} ⩾ 5 y^{4} \Rightarrow 5 y^{5} ⩾ 5 y^{4} z$ $\Rightarrow z ⩽ y$ $3 \Rightarrow z ⩾ x \Rightarrow (1, 2, 3) \Rightarrow x = y = z$ $x \neq 0$ $1 \Leftrightarrow 4 x^{3} = x^{6} + x^{4} + x^{2} + 1 . A M - G M e q u a l i t i e$ $\Rightarrow 1 = x = x^{2} = x^{4} = x^{6}$ $\Rightarrow (x, y, z) = (1, 1, 1)$ $x = 0 \Rightarrow y = 0, z = 0$ $(x, y, z) \in {(1, 1, 1); (0, 0, 0)}$
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