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Question Number 195124 by Shlock last updated on 25/Jul/23

Answered by witcher3 last updated on 25/Jul/23

(√x)+(√y)≤(√((x+1)(y+1))) ⇔x+y+2(√(xy))≤xy+x+y+1⇔xy+1≥2(√(xy)),AM−GM ⇒∀(x,y)∈R_+ (√x)+(√y)≤(√((x+1)(y+1))) ⇒∀(a,b)∈[1,∞[^2 (√(a−1))+(√(b−1))≤(√(ab)) (√(a−1))+(√(b−1))+(√(c−1))≤(√(ab))+(√(c−1))≤(√((ab+1)c))=(√(abc+c)) ⇔(√(a−1))+(√(b−1))+(√(c−1))≤(√(abc+c))

x+y⩽(x+1)(y+1)⇔x+y+2xy⩽xy+x+y+1⇔xy+1⩾2xy,AM−GM⇒∀(x,y)∈R+x+y⩽(x+1)(y+1)⇒∀(a,b)∈[1,∞[2a−1+b−1⩽aba−1+b−1+c−1⩽ab+c−1⩽(ab+1)c=abc+c⇔a−1+b−1+c−1⩽abc+c

Commented by Shlock last updated on 25/Jul/23

Perfect ��

Commented by witcher3 last updated on 26/Jul/23

thank You God bless You

thankYouGodblessYou

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