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Lời giải:
\(\frac{n-1}{n!}=\frac{n}{n!}-\frac{1}{n!}=\frac{1}{(n-1)!}-\frac{1}{n!}\). Do đó:
\(\text{VT}=\frac{1}{1!}-\frac{1}{2!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-....+\frac{1}{(n-1)!}-\frac{1}{n!}=1-\frac{1}{n!}< 1\)
Ta có đpcm.
1)Áp dụng bđt AM-GM:
\(2\left(ab+\frac{a}{b}+\frac{b}{a}\right)=\left(ab+\frac{a}{b}\right)+\left(ab+\frac{b}{a}\right)+\left(\frac{a}{b}+\frac{b}{a}\right)\ge2\left(a+b+1\right)\)
\(\Leftrightarrow ab+\frac{a}{b}+\frac{b}{a}\ge a+b+1."="\Leftrightarrow a=b=1\)
2) Áp dụng bđt AM-GM ta có: \(a+\frac{1}{a-1}=a-1+1+\frac{1}{a-1}\ge2\sqrt{\left(a-1\right).\frac{1}{a-1}}+1=3\)
\("="\Leftrightarrow a=2\)
3) Áp dụng bđt AM-GM:
\(2\left(\frac{bc}{a}+\frac{ac}{b}+\frac{ab}{c}\right)=\left(\frac{ab}{c}+\frac{bc}{a}\right)+\left(\frac{ac}{b}+\frac{ab}{c}\right)+\left(\frac{bc}{a}+\frac{ac}{b}\right)\ge2\left(a+b+c\right)\)
Cộng theo vế và rg => ddpcm. Dấu bằng khi a=b=c
\(2\left(1+abc\right)+\sqrt{2\left(1+a^2\right)\left(1+b^2\right)\left(1+c^2\right)}\)
\(=2\left(1+abc\right)+\sqrt{\left[\left(a+1\right)^2+\left(1-a\right)^2\right]\left[\left(b+c\right)^2+\left(bc-1\right)^2\right]}\)
\(\ge2\left(1+abc\right)+\left(a+1\right)\left(b+c\right)+\left(1-a\right)\left(bc-1\right)\)
\(=\left(1+a\right)\left(1+b\right)\left(1+c\right)\)
\(2\left(1+abc\right)+\sqrt{2\left(1+a^2\right)\left(1+b^2\right)\left(1+c^2\right)}.\)
\(=2\left(1+abc\right)+\sqrt{\left[\left(a+1\right)^2+\left(1-a\right)^2\right]\left[\left(b+c\right)^2+\left(bc-1\right)^2\right]}\)
\(\ge2\left(1+abc\right)+\left(a+1\right)\left(b+c\right)+\left(1-a\right)\left(bc-1\right)\)
\(=\left(1+a\right)\left(1+b\right)\left(1+c\right)\)
\(a^2+b^2+1\ge ab+a+b\)
\(\Leftrightarrow a^2+b^2+1-ab-a-b\ge0\)
\(\Leftrightarrow2a^2+2b^2+2-2ab-2a-2b\ge0\)
\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(a^2-2a+1\right)+\left(b^2-2b+1\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(a-1\right)^2+\left(b-1\right)^2\ge0\)(luôn đúng)
Vậy bđt đã đc chứng minh
Lời giải
Với mọi $n\in\mathbb{N}$ ta có:
\(\frac{1}{\sqrt{1}}> \frac{1}{\sqrt{n}}\)
\(\frac{1}{\sqrt{2}}> \frac{1}{\sqrt{n}}\)
.....
Do đó:
\(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{n}}> \underbrace{\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n}}+...+\frac{1}{\sqrt{n}}}_{\text{n số}}=\frac{n}{\sqrt{n}}=\sqrt{n}\)
(chứng minh xong vế 1)
Vế 2:
\(\frac{1}{2\sqrt{1}}+\frac{1}{2\sqrt{2}}+...+\frac{1}{2\sqrt{n}}< \frac{1}{\sqrt{0}+\sqrt{1}}+\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+...+\frac{1}{\sqrt{n-1}+\sqrt{n}}\)
\(=\frac{\sqrt{1}-\sqrt{0}}{1-0}+\frac{\sqrt{2}-\sqrt{1}}{2-1}+\frac{\sqrt{3}-\sqrt{2}}{3-2}+...+\frac{\sqrt{n}-\sqrt{n-1}}{n-(n-1)}\)
\(=\sqrt{1}-\sqrt{0}+\sqrt{2}-\sqrt{1}+...+\sqrt{n}-\sqrt{n-1}=\sqrt{n}\)
\(\Rightarrow \frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{n}}< 2\sqrt{n}\) (đpcm)
Vậy....