Ta có : \(\dfrac{1}{2^2}< \dfrac{1}{1.2};\dfrac{1}{3^2}< \dfrac{1}{2.3};\dfrac{1}{4^2}< \dfrac{1}{3.4};...;\dfrac{1}{n^2}< \dfrac{1}{\left(n-1\right).n}\)

\(\Rightarrow M=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{n^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{\left(n-1\right).n}\)

\(\Rightarrow M< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{n-1}-\dfrac{1}{n}\)

\(\Rightarrow M< 1-\dfrac{1}{n}< 1\)

Vậy \(M=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{n^2}< 1\)

Để \(M< 1\), ta phải có điều kiện: \(n\in\) R*. Nếu \(n=0\) thì \(M\) không xác định.
\(M=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{n^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{\left(n-1\right)n}\)
                                                 \(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{n-1}-\dfrac{1}{n}\)
                                                 \(=1-\dfrac{1}{n}< 1\)
Vậy \(M< 1\) với \(n\in\) R*.

a, Ta có :

\(M=\dfrac{1}{1\cdot2}+\dfrac{1}{1\cdot2\cdot3}+\dfrac{1}{1\cdot2\cdot3\cdot4}+...+\dfrac{1}{1\cdot2\cdot3\cdot...\cdot100}\\ < \dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{99\cdot100}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-...+\dfrac{1}{99}-\dfrac{1}{100}\\ =1-\dfrac{1}{100}=\dfrac{99}{100}< 1\\ \Rightarrow M< 1\\ \RightarrowĐpcm\)

\(a,M=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{n^2}\)

\(M< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{\left(n-1\right)n}\)

\(M< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{n-1}-\dfrac{1}{n}\)

\(M< 1-\dfrac{1}{n}< 1\)

\(\Rightarrow M< 1\left(đpcm\right)\)

\(b,N=\dfrac{1}{4^2}+\dfrac{1}{6^6}+\dfrac{1}{8^2}+...+\dfrac{1}{\left(2n\right)^2}\)

\(N< \dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+...+\dfrac{1}{\left(2n-1\right)\left(2n+1\right)}\)

\(N< \dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\)

\(N< \dfrac{1}{3}-\dfrac{1}{2n+1}< \dfrac{1}{3}\)

\(c,\) Vì \(a< b\Rightarrow2a< a+b\)

\(c< d\Rightarrow2c< c+d\)

\(m< n\Rightarrow2m< m+n\)

\(\Rightarrow2a+2c+2m=2.\left(a+c+m\right)< a+b+c+d+m+n\)

\(\Rightarrow\dfrac{a+c+m}{a+b+c+d+m}< \dfrac{1}{2}\)

Bài 1)

Áp dụng BĐT Bunhiacopxki ta có:

\(1=(a^2+b^2)(m^2+n^2)\geq (am+bn)^2\Rightarrow -1\leq am+bn\leq 1\)

Dấu bằng xảy ra khi \(\frac{a}{m}=\frac{b}{n}\) . Kết hợp với \(a^2+b^2=m^2+n^2=1\)

\(\Rightarrow \) dấu bằng xảy ra khi \(a=\pm m;b=\pm n\)

Bài 2)

Ta thấy:

\((ac-bd)^2\geq 0\Rightarrow a^2c^2+b^2d^2\geq 2abcd\Rightarrow (ac+bd)^2\geq 4abcd\)

\(\Leftrightarrow 4\geq 4cd\rightarrow cd\leq 1\Rightarrow 1-cd\geq 0\) (đpcm)

Dấu bằng xảy ra khi \(ac=bd=\pm 1\) và \(cd=1\) ....

Bài 3)

Vế đầu:

\(\Leftrightarrow ab+bc+ac\leq a^2+b^2+c^2\)

Nhân $2$ và chuyển vế \(\Leftrightarrow (a-b)^2+(b-c)^2+(c-a)^2\geq 0\)

BĐT trên luôn đúng nên BĐT đầu tiên cũng đúng.

Vế sau:

\(\Leftrightarrow 2(a^2+b^2+c^2)\geq 2(ab+bc+ac)\)

\(\Leftrightarrow (a-b)^2+(b-c)^2+(c-a)^2\geq 0\) (luôn đúng)

Do đó BĐT sau cũng luôn đúng với mọi số thực $a,b,c$

Dấu bằng xảy ra khi $a=b=c$

\(\left\{{}\begin{matrix}m^2+n^2=1\\a^2+b^2=1\end{matrix}\right.\) \(\Leftrightarrow\left(a^2+b^2\right)\left(m^2+n^2\right)=\left(am\right)^2+\left(an\right)^2+\left(bm\right)^2+\left(bn\right)^2=1\)\(\Leftrightarrow\left(am+bn\right)^2-\left[\left(ambn-\left(an\right)^2\right)+\left(ambn-\left(bm\right)^2\right)\right]=1\)\(\Leftrightarrow\left(am+bn\right)^2+\left[an\left(bm-an\right)\right]+\left[bm\left(an-bm\right)\right]=1\)

\(\Leftrightarrow\left(am+bn\right)^2-\left(bm-an\right)\left(an-bm\right)=1\)

\(\Leftrightarrow\left(am+bn\right)^2+\left(an-bm\right)^2=1\\ \)

\(\left(an-bm\right)^2\ge0\forall_{a,b,m,n}\Rightarrow\left(am+bn\right)^2\le1\)

\(\Rightarrow-1\le\left(am+bn\right)\le1\Rightarrow dpcm\)

nhanh giúp mình