\(2A=\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+\dfrac{2}{3.4.5}+...+\dfrac{2}{18.19.20}\)

\(\Rightarrow2A=\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+...+\dfrac{1}{18.19}-\dfrac{1}{19.20}\)

\(\Rightarrow2A=\dfrac{1}{1.2}-\dfrac{1}{19.20}< \dfrac{1}{1.2}\)

\(\Rightarrow2A< \dfrac{1}{2}\)

\(\Rightarrow A< \dfrac{1}{4}\) (đpcm)

lỗi

lỗi cực kỳ

Đây này: chứng minh $(a^m-1, a^n-1)=a^{(m,n)}-1$ - Số học - Diễn đàn Toán học

Chưa lên mạng thì hỏi làm gì?

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\(M=\left(\dfrac{1}{2+2\sqrt{a}}+\dfrac{1}{2-2\sqrt{a}}-\dfrac{a^2+1}{1-a^2}\right)\)

\(=\left(\dfrac{1}{2\left(1+\sqrt{a}\right)}+\dfrac{1}{2\left(1-\sqrt{a}\right)}-\dfrac{a^2+1}{1-a^2}\right).\dfrac{1+a}{a}\)

\(=\left(\dfrac{1-\sqrt{a}+1+\sqrt{a}}{2\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}-\dfrac{a^2+1}{\left(1-a\right)\left(1+a\right)}\right).\dfrac{1+a}{a}\)

\(=\left(\dfrac{1}{1-a}-\dfrac{a^2+1}{\left(1-a\right)\left(1+a\right)}\right).\dfrac{1+a}{a}\)

\(=\dfrac{1+a-a^2-1}{\left(1-a\right)\left(1+a\right)}.\dfrac{1+a}{a}\) (nghĩa là 1+a - (a^2 + 1 ) phá ngoặc thì đổi dấu như kia nhé.

quên mk chưa lm xong đã gửi r

\(=\dfrac{-a^2+a}{\left(1-a\right)\left(1+a\right)}.\dfrac{1+a}{a}\)

\(\dfrac{a\left(1-a\right)}{\left(1-a\right)\left(1+a\right)}.\dfrac{1+a}{a}=1\)( chia hết cho nhau thì = 1 nhé

Áp dụng BĐT AM-GM ta có:

\(\dfrac{a^3}{\left(1+b\right)\left(1+c\right)}+\dfrac{b+1}{8}+\dfrac{c+1}{8}\)

\(\ge3\sqrt[3]{\dfrac{a^3}{\left(1+b\right)\left(1+c\right)}\cdot\dfrac{b+1}{8}\cdot\dfrac{c+1}{8}}=\dfrac{3a}{4}\)

Tương tự cho 2 BĐT còn lại ta cũng có:

\(\dfrac{b^3}{\left(1+c\right)\left(1+a\right)}+\dfrac{c+1}{8}+\dfrac{a+1}{8}\ge\dfrac{3b}{4};\dfrac{c^3}{\left(1+a\right)\left(1+b\right)}+\dfrac{a+1}{8}+\dfrac{b+1}{8}\ge\dfrac{3c}{4}\)

Cộng theo vế 3 BĐT trên ta có:

\(VT+\dfrac{2\left(a+b+c+3\right)}{8}\ge\dfrac{3\left(a+b+c\right)}{4}\)

\(\Leftrightarrow VT+\dfrac{2\left(3\sqrt[3]{abc}+3\right)}{8}\ge\dfrac{3\cdot3\sqrt[3]{abc}}{4}\Leftrightarrow VT\ge\dfrac{3}{4}=VP\)

Khi \(a=b=c=1\)

khó

Đặt \(\left\{{}\begin{matrix}x=\dfrac{1}{a}\\y=\dfrac{1}{b}\\z=\dfrac{1}{c}\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x,y,z>0\\xyz=1\end{matrix}\right.\) và BĐT cần chứng minh là:

\(\dfrac{x^2}{y+z}+\dfrac{y^2}{x+z}+\dfrac{z^2}{x+y}\ge\dfrac{3}{2}\)

Áp dụng BĐT Cauchy-Schwarz dạng Engel và AM-GM ta có:

\(VT=\dfrac{x^2}{y+z}+\dfrac{y^2}{x+z}+\dfrac{z^2}{x+y}\)

\(\ge\dfrac{\left(x+y+z\right)^2}{2\left(x+y+z\right)}=\dfrac{x+y+z}{2}\ge\dfrac{3\sqrt[3]{xyz}}{2}=\dfrac{3}{2}=VP\)

Xảy ra khi \(x=y=z=1 \Rightarrow a=b=c=1\)

ai tick cho mik , mik tick lại cho !^__<nhớ giải câu hỏi nhé ! thanks