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Question Number 109178 by Algoritm last updated on 21/Aug/20

Commented by Algoritm last updated on 21/Aug/20

$mr W$
	Answered by Dwaipayan Shikari last updated on 21/Aug/20

	Commented by Algoritm last updated on 21/Aug/20

	$? ? ?$
	Commented by aurpeyz last updated on 21/Aug/20

	$w o w . H o w d o y o u g e t t o t h i n k o f t h i s p r o c e s s ?$ $i l i k e i t$
	Commented by floor(10²Eta[1]) last updated on 22/Aug/20

	$he just used the AM - HM inequality and$ $the AM - GM inequality$
	Answered by 1549442205PVT last updated on 22/Aug/20
	$We have: (x+y+z)((1/x)+(1/y)+(1/z))+((8xyz)/((x+y)(y+z)(z+x)))≥10 ⇔(((x+y+z)(xy+yz+zx))/(xyz))+((8xyz)/((x+y)(y+z)(z+x)))≥10(1) First,we note some following identities: (x+y)(y+z)(z+x)=(x+y+z)(xy+yz+zx)−xyz Hence,The inequality (1) is equivalent to (((x+y+z)(xy+yz+zx))/(xyz))+((8xyz)/((x+y+z)(xy+yz+zx)−xyz))≥10(2) Since the nominator and denomirator of each the fraction in the LHS are in same degree (degree 3),WLOG we can suppose that x+y+z=3(because if x+y+z=m then put x=((am)/3),y=((bm)/3), z=((cm)/3)⇒((am)/3)+((bm)/3)+((cm)/3)=m⇒a+b+c=3 and the inequality (2) becomes (((m/3)^3 (a+b+c)(ab+bc+ca))/((m/3)^3 abc))+((8(m/3)^3 abc)/((m/3)^3 [(a+b+c)(ab+bc+ca)−abc]))≥10 ⇔(( (a+b+c)(ab+bc+ca))/(abc))+((8abc)/((a+b+c)(ab+bc+ca)−abc))≥10 which is also the inequality (2) but under the condition a+b+c=3) By that reason we have (2)⇔((3(xy+yz+zx))/(xyz))+((8xyz)/(3(xy+yz+zx)−xyz))≥10 ⇔9(xy+yz+zx)^2 −3xyz(xy+yz+zx) +18(xyz)^2 ≥10xyz[3(xy+yz+zx)−xyz] ⇔9(xy+yz+zx)^2 +18(xyz)^2 ≥ 33xyz(xy+yz+zx)(3) Now we put xy=m,xz=n,yz=p.Then 3^2 =(x+y+z)^2 ≥3(xy+yz+zx) ⇒m+n+p=xy+yz+zx≤3 3≥xy+yz+zx≥3^3 (√(xy.yz.zx))=3^3 (√((xyz)^2 )) ⇒^3 (√(xyz)) ≤1,mnp≤(((m+m+p)/3))^3 ≤1(∗) We have also (xyz)^2 =mnp.Therefore, (3)⇔9(m+n+p)^2 +18mnp≥33(m+n+p)(√(mnp)) (3′) Apply Cauchy′s inequality for two positive numbers we have 2(m+n+p)^2 +18mnp≥(√(2(m+n+p)^2 .8mnp)) =8(m+n+p)(√(mnp)) (i).On the other hands, Since mnp≤1⇒(√(mnp)) ≤1(due to (∗)) ⇒mnp(√(mnp))=(√((mnp)^3 ))≤mnp ⇒(√(mnp))≤^3 (√(mnp)) (∗∗).Now apply Cauchy′s inequality for three positive numbers we get: m^2 +mn+mp≥3^3 (√(m^2 .mn.mp)) =3m^3 (√(mnp)) ≥3(√(mnp)) (due to (∗∗)).Similarly, n^2 +nm+np≥3n(√(mnp)) p^2 +pn+pm≥3p(√(mnp)) Adding up three above inequalities we get m^2 +n^2 +p^2 +2(mn+mp+np) =(m+n+p)^2 ≥3(m+n+p)(√(mnp)) ⇒7(m+n+p)^3 ≥21(m+n+p)(√(mnp))(ii) Adding up two inequalities (i)and(ii) we get: 9(m+n+p)^2 +18mnp≥33(m+n+p)(√(mnp)) ,so the inequality (3′)is true which implies that (2) is true .Consequently, the inequality (1)is proved.Thus,we proved the inequality: (x+y+z)((1/x)+(1/y)+(1/z))+((8xyz)/((x+y)(y+z)(z+x)))≥10 The equality ocurrs if and only if x=y=z (Q.E.D)$
	$We have :$ $(x + y + z) (\frac{1}{x} + \frac{1}{y} + \frac{1}{z}) + \frac{8 xyz}{(x + y) (y + z) (z + x)} ⩾ 10$ $\Leftrightarrow \frac{(x + y + z) (xy + yz + zx)}{xyz} + \frac{8 xyz}{(x + y) (y + z) (z + x)} ⩾ 10 (1)$ $First, we note some following identities :$ $(x + y) (y + z) (z + x) = (x + y + z) (xy + yz + zx) - xyz$ $Hence, The inequality (1) is equivalent to$ $\frac{(x + y + z) (xy + yz + zx)}{xyz} + \frac{8 xyz}{(x + y + z) (xy + yz + zx) - xyz} ⩾ 10 (2)$ $Since the nominator and denomirator$ $of each the fraction in the LHS are$ $in same degree (degree 3), WLOG we$ $can suppose that x + y + z = 3 (because$ $if x + y + z = m then put x = \frac{am}{3}, y = \frac{bm}{3},$ $z = \frac{cm}{3} \Rightarrow \frac{am}{3} + \frac{bm}{3} + \frac{cm}{3} = m \Rightarrow a + b + c = 3$ $and the inequality (2) becomes$ $\frac{{(m / 3)}^{3} (a + b + c) (ab + bc + ca)}{{(m / 3)}^{3} abc} + \frac{8 {(m / 3)}^{3} abc}{{(m / 3)}^{3} [(a + b + c) (ab + bc + ca) - abc]} ⩾ 10$ $\Leftrightarrow \frac{(a + b + c) (ab + bc + ca)}{abc} + \frac{8 abc}{(a + b + c) (ab + bc + ca) - abc} ⩾ 10$ $which is also the inequality (2) but$ $under the condition a + b + c = 3)$ $By that reason we have$ $(2) \Leftrightarrow \frac{3 (xy + yz + zx)}{xyz} + \frac{8 xyz}{3 (xy + yz + zx) - xyz} ⩾ 10$ $\Leftrightarrow 9 {(xy + yz + zx)}^{2} - 3 xyz (xy + yz + zx)$ $+ 18 {(xyz)}^{2} ⩾ 10 xyz [3 (xy + yz + zx) - xyz]$ $\Leftrightarrow 9 {(xy + yz + zx)}^{2} + 18 {(xyz)}^{2} ⩾$ $33 xyz (xy + yz + zx) (3)$ $Now we put xy = m, xz = n, yz = p . Then$ $3^{2} = {(x + y + z)}^{2} ⩾ 3 (xy + yz + zx)$ $\Rightarrow m + n + p = xy + yz + zx ⩽ 3$ $3 ⩾ xy + yz + zx ⩾ 3^{3} \sqrt{xy . yz . zx} = 3^{3} \sqrt{{(xyz)}^{2}}$ $\Rightarrow^{3} \sqrt{xyz} ⩽ 1, mnp ⩽ {(\frac{m + m + p}{3})}^{3} ⩽ 1 ()$ $We have also {(xyz)}^{2} = mnp . Therefore,$ $(3) \Leftrightarrow 9 {(m + n + p)}^{2} + 18 mnp ⩾ 33 (m + n + p) \sqrt{mnp} (3^{'})$ $Apply {Cauchy}^{'} s inequality for two$ $positive numbers we have$ $2 {(m + n + p)}^{2} + 18 mnp ⩾ \sqrt{2 {(m + n + p)}^{2} . 8 mnp}$ $= 8 (m + n + p) \sqrt{mnp} (i) . On the other hands,$ $Since mnp ⩽ 1 \Rightarrow \sqrt{mnp} ⩽ 1 (due to ())$ $\Rightarrow mnp \sqrt{mnp} = \sqrt{{(mnp)}^{3}} ⩽ mnp$ $\Rightarrow \sqrt{mnp} ⩽^{3} \sqrt{mnp} (* ) . Now apply$ ${Cauchy}^{'} s inequality for three positive$ $numbers we get :$ $m^{2} + mn + mp ⩾ 3^{3} \sqrt{m^{2} . mn . mp} = 3 m^{3} \sqrt{mnp}$ $⩾ 3 \sqrt{mnp} (due to ( *)) . Similarly,$ $n^{2} + nm + np ⩾ 3 n \sqrt{mnp}$ $p^{2} + pn + pm ⩾ 3 p \sqrt{mnp}$ $Adding up three above inequalities$ $we get m^{2} + n^{2} + p^{2} + 2 (mn + mp + np)$ $= {(m + n + p)}^{2} ⩾ 3 (m + n + p) \sqrt{mnp}$ $\Rightarrow 7 {(m + n + p)}^{3} ⩾ 21 (m + n + p) \sqrt{mnp} (ii)$ $Adding up two inequalities (i) and (ii)$ $we get :$ $9 {(m + n + p)}^{2} + 18 mnp ⩾ 33 (m + n + p) \sqrt{mnp}$ $, so the inequality (3^{'}) is true which$ $implies that (2) is true . Consequently,$ $the inequality (1) is proved . Thus, we$ $proved the inequality :$ $(x + y + z) (\frac{1}{x} + \frac{1}{y} + \frac{1}{z}) + \frac{8 xyz}{(x + y) (y + z) (z + x)} ⩾ 10$ $The equality ocurrs if and only if$ $x = y = z (Q . E . D)$
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