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Question Number 144691 by mnjuly1970 last updated on 27/Jun/21

	Answered by mnjuly1970 last updated on 27/Jun/21
	$ans: a ,b , c , x , y , z positive real number : titus lemma: (a^2 /x^( 2 ) ) +(b^2 /y^( 2) )+(c^( 2) /z^( 2) ) ≥(((a+b+c)^( 2) )/(x+y+x)) ....✓ 1:= (1/x)+(1/y)+(1/z) = (1^2 /x)+(1^2 /y)+(1^2 /z)≥(((1+1+1)^2 )/(x+y+z)) =(9/(x+y+z)) ∴ x+y +z ≥ 9 ....(∗) M:= (x−1)(y−1)(z−1) := (xy−x−y+1)(z−1) :=[{ xyz −xy−xz−yz}=^? 0]+x+y+z−1 :=x+y+z−1≥9−1 =8 ....✓ ( ? ): (1/x)+(1/y)+(1/z)=1 ⇒ xy+yz+zx−xyz=0$
	$a n s :$ $a, b, c, x, y, z p o s i t i v e r e a l$ $n u m b e r :$ $t i t u s l e m m a :$ $\frac{a^{2}}{x^{2}} + \frac{b^{2}}{y^{2}} + \frac{c^{2}}{z^{2}} ⩾ \frac{{(a + b + c)}^{2}}{x + y + x} . . . . ✓$ $1 := \frac{1}{x} + \frac{1}{y} + \frac{1}{z}$ $= \frac{1^{2}}{x} + \frac{1^{2}}{y} + \frac{1^{2}}{z} ⩾ \frac{{(1 + 1 + 1)}^{2}}{x + y + z} = \frac{9}{x + y + z}$ $∴ x + y + z ⩾ 9 . . . . (*)$ $M := (x - 1) (y - 1) (z - 1)$ $:= (x y - x - y + 1) (z - 1)$ $:= [{x y z - x y - x z - y z} \overset{?}{=} 0] + x + y + z - 1$ $:= x + y + z - 1 ⩾ 9 - 1 = 8 . . . . ✓$ $(?) : \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 1 \Rightarrow x y + y z + z x - x y z = 0$
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