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Question Number 8259 by 314159 last updated on 04/Oct/16

Answered by Yozzias last updated on 04/Oct/16

Let u(x)=(1+x)^n (n∈N,n≥2) 2^n −1=C_1 +C_2 +C_3 +C_4 +...+C_n (2^n −1)^(1/2) =(C_1 +C_2 +C_3 +C_4 +...+C_n )^(1/2) The Root Mean Square Inequality states that for x_i >0 (i=1,2,3,...,n), (√((Σ_(i=1) ^n x_i ^2 )/n))≥((Σ_(i=1) ^n x_i )/n). Let x_i =(√C_i ) . (√((Σ_(i=1) ^n ((√C_i ))^2 )/n))≥((Σ_(i=1) ^n (√C_i ))/n) (((C_1 +C_2 +C_3 +...+C_n )/n))^(1/2) ≥(((√C_1 )+(√C_2 )+(√C_3 )+...+(√C_n ))/n) ×n⇒(n(C_1 +C_2 +C_3 +...+C_n ))^(1/2) ≥(√C_1 )+(√C_2 )+(√C_3 )+...+(√C_n ) But (2^n −1)^(1/2) =(C_1 +C_2 +C_3 +C_4 +...+C_n )^(1/2) . ∴ (n(2^n −1))^(1/2) ≥(√C_1 )+(√C_2 )+(√C_3 )+...+(√C_n ).

$$\mathrm{Let}\:\mathrm{u}\left(\mathrm{x}\right)=\left(\mathrm{1}+\mathrm{x}\right)^{\mathrm{n}} \:\:\left(\mathrm{n}\in\mathbb{N},\mathrm{n}\geqslant\mathrm{2}\right) \\ $$$$\mathrm{2}^{\mathrm{n}} −\mathrm{1}=\mathrm{C}_{\mathrm{1}} +\mathrm{C}_{\mathrm{2}} +\mathrm{C}_{\mathrm{3}} +\mathrm{C}_{\mathrm{4}} +...+\mathrm{C}_{\mathrm{n}} \\ $$$$\left(\mathrm{2}^{\mathrm{n}} −\mathrm{1}\right)^{\mathrm{1}/\mathrm{2}} =\left(\mathrm{C}_{\mathrm{1}} +\mathrm{C}_{\mathrm{2}} +\mathrm{C}_{\mathrm{3}} +\mathrm{C}_{\mathrm{4}} +...+\mathrm{C}_{\mathrm{n}} \right)^{\mathrm{1}/\mathrm{2}} \\ $$$$\mathrm{The}\:\mathrm{Root}\:\mathrm{Mean}\:\mathrm{Square}\:\mathrm{Inequality} \\ $$$$\mathrm{states}\:\mathrm{that}\:\mathrm{for}\:\mathrm{x}_{\mathrm{i}} >\mathrm{0}\:\left(\mathrm{i}=\mathrm{1},\mathrm{2},\mathrm{3},...,\mathrm{n}\right), \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\sqrt{\frac{\underset{\mathrm{i}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\mathrm{x}_{\mathrm{i}} ^{\mathrm{2}} }{\mathrm{n}}}\geqslant\frac{\underset{\mathrm{i}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\mathrm{x}_{\mathrm{i}} }{\mathrm{n}}. \\ $$$$\mathrm{Let}\:\mathrm{x}_{\mathrm{i}} =\sqrt{\mathrm{C}_{\mathrm{i}} }\:. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\sqrt{\frac{\underset{\mathrm{i}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\left(\sqrt{\mathrm{C}_{\mathrm{i}} }\right)^{\mathrm{2}} }{\mathrm{n}}}\geqslant\frac{\underset{\mathrm{i}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\sqrt{\mathrm{C}_{\mathrm{i}} }}{\mathrm{n}} \\ $$$$\left(\frac{\mathrm{C}_{\mathrm{1}} +\mathrm{C}_{\mathrm{2}} +\mathrm{C}_{\mathrm{3}} +...+\mathrm{C}_{\mathrm{n}} }{\mathrm{n}}\right)^{\mathrm{1}/\mathrm{2}} \geqslant\frac{\sqrt{\mathrm{C}_{\mathrm{1}} }+\sqrt{\mathrm{C}_{\mathrm{2}} }+\sqrt{\mathrm{C}_{\mathrm{3}} }+...+\sqrt{\mathrm{C}_{\mathrm{n}} }}{\mathrm{n}} \\ $$$$×\mathrm{n}\Rightarrow\left(\mathrm{n}\left(\mathrm{C}_{\mathrm{1}} +\mathrm{C}_{\mathrm{2}} +\mathrm{C}_{\mathrm{3}} +...+\mathrm{C}_{\mathrm{n}} \right)\right)^{\mathrm{1}/\mathrm{2}} \geqslant\sqrt{\mathrm{C}_{\mathrm{1}} }+\sqrt{\mathrm{C}_{\mathrm{2}} }+\sqrt{\mathrm{C}_{\mathrm{3}} }+...+\sqrt{\mathrm{C}_{\mathrm{n}} } \\ $$$$\mathrm{But}\:\left(\mathrm{2}^{\mathrm{n}} −\mathrm{1}\right)^{\mathrm{1}/\mathrm{2}} =\left(\mathrm{C}_{\mathrm{1}} +\mathrm{C}_{\mathrm{2}} +\mathrm{C}_{\mathrm{3}} +\mathrm{C}_{\mathrm{4}} +...+\mathrm{C}_{\mathrm{n}} \right)^{\mathrm{1}/\mathrm{2}} . \\ $$$$\therefore\:\left(\mathrm{n}\left(\mathrm{2}^{\mathrm{n}} −\mathrm{1}\right)\right)^{\mathrm{1}/\mathrm{2}} \geqslant\sqrt{\mathrm{C}_{\mathrm{1}} }+\sqrt{\mathrm{C}_{\mathrm{2}} }+\sqrt{\mathrm{C}_{\mathrm{3}} }+...+\sqrt{\mathrm{C}_{\mathrm{n}} }. \\ $$$$ \\ $$$$ \\ $$

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