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Question Number 197464 by universe last updated on 18/Sep/23

	Answered by witcher3 last updated on 19/Sep/23
	$claim (((a+b)^6 )/((ab)^2 ))≥32(a^2 +b^2 )....P,by symetri a≥b a=tb,t∈]0,1] (P)⇔∀t∈[0,1[(1+t)^6 ≥32t^2 (1+t^2 ) ⇔t^6 +6t^5 −17t^4 +20t^3 −17t^2 +6t+1≥0 ⇔t^3 +(1/t^3 )+6(t^2 +(1/t^2 ))−17(t+(1/t))+20 y=t+(1/t)⇒y∈[2,∞[=I ⇔∀y∈I y^3 +6y^2 −20y+8≥0 (y−2)(y^2 +8y−4)≥0 True y>2 y^2 +8y−4=y^2 −4+8y≥4−4+8y≥16 True⇒(P) True ⇒∀(a,b)∈R_+ (((a_1 +a_2 )^6 )/((a_1 a_2 )^4 ))≥32(a_1 ^2 +a_2 ^2 )^2 Σ_(cyc) (((a_1 +a_2 )^6 )/((a_1 a_2 )^4 ))≥32Σ_(cyc) (a_1 ^2 +a_2 ^2 ) S={a_i ,i∈[1,n]},a_(n+1) =a_1 ⇒Σ_(i=1) ^n (((a_i +a_(i+1) )^2 )/((a_i a_(i+1) )^2 ))≥32Σ_(i=1) ^(n+1) (a_i ^2 +a_(i+1) ^2 )=32.2Σ_(i=1) ^n a_i ^2 Σ_(cyc) (((a_1 +a_2 )^6 )/(a_1 ^2 a_2 ^2 ))≥64Σ_(i=1) ^n a_i ^2$
	$claim$ $\frac{{(a + b)}^{6}}{{(ab)}^{2}} ⩾ 32 (a^{2} + b^{2}) . . . . P, by symetri a ⩾ b$ $a = tb, t \in] 0, 1]$ $(P) \Leftrightarrow \forall t \in [0, 1 [{(1 + t)}^{6} ⩾ {32 t}^{2} (1 + t^{2})$ $\Leftrightarrow t^{6} + {6 t}^{5} - {17 t}^{4} + {20 t}^{3} - {17 t}^{2} + 6 t + 1 ⩾ 0$ $\Leftrightarrow t^{3} + \frac{1}{t^{3}} + 6 (t^{2} + \frac{1}{t^{2}}) - 17 (t + \frac{1}{t}) + 20$ $y = t + \frac{1}{t} \Rightarrow y \in [2, \infty [= I$ $\Leftrightarrow \forall y \in I y^{3} + {6 y}^{2} - 20 y + 8 ⩾ 0$ $(y - 2) (y^{2} + 8 y - 4) ⩾ 0$ $True y > 2$ $y^{2} + 8 y - 4 = y^{2} - 4 + 8 y ⩾ 4 - 4 + 8 y ⩾ 16$ $True \Rightarrow (P) True$ $\Rightarrow \forall (a, b) \in R_{+} \frac{{(a_{1} + a_{2})}^{6}}{{(a_{1} a_{2})}^{4}} ⩾ 32 {(a_{1}^{2} + a_{2}^{2})}^{2}$ $\sum_{cyc} \frac{{(a_{1} + a_{2})}^{6}}{{(a_{1} a_{2})}^{4}} ⩾ 32 \sum_{cyc} (a_{1}^{2} + a_{2}^{2})$ $S = {a_{i}, i \in [1, n]}, a_{n + 1} = a_{1}$ $\Rightarrow \underset{i = 1}{\sum^{n}} \frac{{(a_{i} + a_{i + 1})}^{2}}{{(a_{i} a_{i + 1})}^{2}} ⩾ 32 \underset{i = 1}{\sum^{n + 1}} (a_{i}^{2} + a_{i + 1}^{2}) = 32.2 \underset{i = 1}{\sum^{n}} a_{i}^{2}$ $\sum_{cyc} \frac{{(a_{1} + a_{2})}^{6}}{a_{1}^{2} a_{2}^{2}} ⩾ 64 \underset{i = 1}{\sum^{n}} a_{i}^{2}$
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