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E.202. Problema săptămânii 258

Dacă a,b,c>1a, b, c > 1 satisfac 1a−1+1b−1+1c−1=1,\dfrac{1}{a-1} + \dfrac{1}{b-1} + \dfrac{1}{c-1} = 1, demonstrați că abc≥64.abc \geq 64.

Soluție
Soluție:

Facem substituțiile x=1a−1,y=1b−1,z=1c−1x=\dfrac{1}{a-1}, y=\dfrac{1}{b-1}, z=\dfrac{1}{c-1} și obținem enunțul echivalent:

x,y,z>0,x+y+z=1⇒(1+1x)(1+1y)(1+1z)≥64.x, y, z > 0, x+y+z=1 \Rightarrow \Big(1+\dfrac{1}{x}\Big)\Big(1+\dfrac{1}{y}\Big)\Big(1+\dfrac{1}{z}\Big) \geq 64.

Vom demonstra această nouă inegalitate:

(1+1x)(1+1y)(1+1z)≥Huygens∗(1+1x⋅1y⋅1z3)3≥hm−gm(1+3x+y+z)3=(1+31)3=64.\Big(1+\dfrac{1}{x}\Big)\Big(1+\dfrac{1}{y}\Big)\Big(1+\dfrac{1}{z}\Big) \overset{Huygens^*}{\geq} \Big(1+\sqrt[3]{\dfrac{1}{x} \cdot \dfrac{1}{y} \cdot \dfrac{1}{z}} \Big)^3 \overset{hm-gm}{\geq} \Big( 1+\dfrac{3}{x+y+z} \Big)^3 =\Big( 1+ \dfrac{3}{1} \Big)^3 = 64.

(*) Inegalitatea lui Huygens: (1+a1)(1+a2)…(1+an)≥(1+a1a2…ann)n,ai≥0.(1+a_1)(1+a_2) \dots(1+a_n) \geq(1+\sqrt[n]{a_1 a_2 \dots a_n})^n, a_i \geq 0.

E.214. Fie a,b∈(0,1)a,b \in (0,1) astfel încât a2+b2=1.a^2+b^2=1. Demonstrați că a+b+1a+1b≥32.a+b+\dfrac{1}{a} + \dfrac{1}{b} \geq 3\sqrt{2}.

MM, 29.02.2024, clasa a 9-a.
Soluție
Soluție:

Soluția 1 (LM). Notăm a=sin⁡x,a=\sin{x}, b=cos⁡x,b=\cos{x}, cu x∈(0,π2).x\in\Big(0,\dfrac{\pi}{2}\Big). Deci și sin⁡2x∈(0,1).\boxed{\sin{2x} \in (0,1)}.

a+b+1a+1b=sin⁡x+cos⁡x+1sin⁡x+1cos⁡x≥Am−Gma+b+\dfrac{1}{a} + \dfrac{1}{b} = \sin{x} + \cos{x} + \dfrac{1}{\sin{x}} + \dfrac{1}{\cos{x}} \overset{Am-Gm}{\geq}

≥2sin⁡xcos⁡x+2sin⁡xcos⁡x=2⋅sin⁡2x+22sin⁡2x=\geq 2\sqrt{\sin{x}\cos{x}} + \dfrac{2}{\sqrt{\sin{x}\cos{x}}} =\sqrt{2} \cdot \sqrt{\sin{2x}} + \dfrac{2\sqrt{2}}{\sqrt{\sin{2x}}}=

=2((sin⁡2x−1⏞≤0)(sin⁡2x−2⏞≤0)sin⁡2x+3)≥32.=\sqrt{2}\Big(\dfrac{(\overbrace{\sqrt{\sin{2x}}-1}^{\leq 0})(\overbrace{\sqrt{\sin{2x}}-2}^{\leq 0})}{\sqrt{\sin{2x}}} + 3 \Big)\geq 3\sqrt{2}.

Soluția 2 (Filip Munteanu). Notăm a+b=s.\boxed{a+b=s}.

a+b2≤a2+b22=12⇒s≤2.\dfrac{a+b}{2} \leq \sqrt{\dfrac{a^2+b^2}{2}}=\dfrac{1}{\sqrt{2}} \Rightarrow \boxed{s \leq \sqrt{2}}.

a+b+1a+1b≥a+b+4a+b.a+b+\dfrac{1}{a} + \dfrac{1}{b} \geq a+b+\dfrac{4}{a+b}. Deci problema se reduce la a demonstra inegalitatea s2−32s+4≥0.s^2-3\sqrt{2}s+4\geq 0.

Soluția 3 (Filip Munteanu). Funcția f:(0,1)→R,f:(0,1)\rightarrow \R, f(x)=x+1xf(x)=x+\dfrac{1}{x} este convexă și descrescătoare, deci
f(a)+f(b)≥2⋅f(a+b2)≥2⋅f(22)=32.f(a)+f(b) \geq 2 \cdot f\Big(\dfrac{a+b}{2}\Big) \geq 2 \cdot f\Big(\dfrac{\sqrt{2}}{2}\Big) = 3\sqrt{2}.

Soluția 4 (Alin Pop). 1=a2+b2≥2ab⇒1ab≥2.1=a^2+b^2 \geq 2ab \Rightarrow \boxed{\dfrac{1}{ab} \geq 2}.

a+1a=a+12a+12a≥314a3.a+\dfrac{1}{a} = a+\dfrac{1}{2a} + \dfrac{1}{2a} \geq 3\sqrt[3]{\dfrac{1}{4a}}. Analog, b+1b≥314b3.b+\dfrac{1}{b} \geq 3\sqrt[3]{\dfrac{1}{4b}}.

a+b+1a+1b≥3(14a3+14b3)≥6116ab6≥62166=32.a+b+\dfrac{1}{a} + \dfrac{1}{b} \geq 3\Big(\sqrt[3]{\dfrac{1}{4a}}+\sqrt[3]{\dfrac{1}{4b}} \Big) \geq 6\sqrt[6]{\dfrac{1}{16ab}} \geq 6\sqrt[6]{\dfrac{2}{16}} = 3\sqrt{2}.