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Ta có :
\(\dfrac{1}{1+a^2}+\dfrac{1}{1+b^2}\ge\dfrac{2}{1+ab}\)
\(\Leftrightarrow\) \(\dfrac{1}{1+a^2}-\dfrac{1}{1+ab}+\dfrac{1}{1+b^2}-\dfrac{1}{1+ab}\ge0\)
\(\Leftrightarrow\) \(\dfrac{1+ab-1-a^2}{\left(1+a^2\right)\left(1+ab\right)}+\dfrac{1+ab-1-b^2}{\left(1+b^2\right)\left(1+ab\right)}\ge0\)
\(\Leftrightarrow\) \(\dfrac{a\left(b-a\right)}{\left(1+a^2\right)\left(1+ab\right)}+\dfrac{b\left(a-b\right)}{\left(1+b^2\right)\left(1+ab\right)}\ge0\)
\(\Leftrightarrow a\left(b-a\right)\left(1+b^2\right)+b\left(a-b\right)\left(1+a^2\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)\left[-a\left(1+b^2\right)+b\left(1+a^2\right)\right]\ge0\)
\(\Leftrightarrow\left(a-b\right)\left(-a-ab^2+b+a^2b\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)\left[ab\left(a-b\right)-\left(a-b\right)\right]\ge0\)
\(\Leftrightarrow\left(a-b\right)\left(a-b\right)\left(ab-1\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)^2\left(ab-1\right)\ge0\) (*)
Vì \(a.b=1\Rightarrow ab-1=0,\left(a-b\right)^2\ge0\)
Do đó (*) đúng . Vậy \(\dfrac{1}{1+a^2}+\dfrac{1}{1+b^2}\ge\dfrac{2}{1+ab}\left(đpcm\right)\)
Áp dụng bất đẳng thức AM-GM ta có:
\(ab+\dfrac{1}{ab}\ge2\sqrt{ab.\dfrac{1}{ab}}\)
\(\Rightarrow ab+\dfrac{1}{ab}\ge2.\sqrt{1}=2.1=2\)
Dâu "=" sảy ra khi và chỉ khi \(a=b=1\)
Vậy GTNN của biểu thức là 2 đạt được khi và chỉ khi \(a=b=1\)
Chúc bạn học tốt!!!
Áp dụng bđt AM-GM ta có:
\(1\ge a+b\ge2\sqrt{ab}\) \(\Leftrightarrow1\ge4ab\)\(\Leftrightarrow\dfrac{1}{4}\ge ab\)
\(S=ab+\dfrac{1}{ab}=ab+\dfrac{1}{16ab}+\dfrac{15}{16ab}\ge2\sqrt{ab.\dfrac{1}{16ab}}+\dfrac{15}{16ab}\) \(\Leftrightarrow S\ge2.\dfrac{1}{4}+\dfrac{15}{16ab}=\dfrac{1}{2}+\dfrac{15}{16ab}\ge\dfrac{1}{2}+\dfrac{15}{16.\dfrac{1}{4}}=\dfrac{17}{4}\)
Dấu "=" xảy ra \(\Leftrightarrow a=b=\dfrac{1}{2}\)
a) Đặt \(A=\dfrac{1}{2^2}+\dfrac{1}{4^2}+\dfrac{1}{6^2}+...+\dfrac{1}{\left(2n\right)^2}\)
\(A=\dfrac{1}{2^2}\left(1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{n^2}\right)\)
Ta có:
\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{n^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{\left(n-1\right)n}\)
\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{n^2}< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{n-1}-\dfrac{1}{n}\)
\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{n^2}< 1-\dfrac{1}{n}\)
\(\Rightarrow1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{n^2}< 1-\dfrac{1}{n}+1\)
\(\Rightarrow1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{n^2}< 2-\dfrac{1}{n}\)
\(\Rightarrow\dfrac{1}{2^2}\left(1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{n^2}\right)< \dfrac{1}{2^2}\left(2-\dfrac{1}{2}\right)\)
\(\Rightarrow A< \dfrac{1}{2^2}.2-\dfrac{1}{2^2}.\dfrac{1}{2}\)
\(\Rightarrow A< \dfrac{1}{2}-\dfrac{1}{2^3}< \dfrac{1}{2}\)
Vậy \(A< \dfrac{1}{2}\left(Đpcm\right)\)
b) Đặt \(B=\dfrac{1}{3^2}+\dfrac{1}{5^2}+\dfrac{1}{7^2}+...+\dfrac{1}{\left(2n+1\right)^2}\)
Ta có:
\(B< \dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{\left(2n-1\right)\left(2n+1\right)}\)
\(B< \dfrac{1}{2}\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{\left(2n-1\right)\left(2n+1\right)}\right)\)
\(B< \dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\right)\)
\(B< \dfrac{1}{2}\left(1-\dfrac{1}{2n+1}\right)\)
\(B< \dfrac{1}{2}\left(\dfrac{2n+1}{2n+1}-\dfrac{1}{2n+1}\right)\)
\(B< \dfrac{1}{2}.\dfrac{2n}{2n+1}\)
\(B< \dfrac{2n}{4n+2}\)
\(B< \dfrac{2n}{2\left(2n+1\right)}\)
\(B< \dfrac{n}{2n+1}\)
CM cái sau:
Ta có: \(a+\frac{1}{a}=\frac{a}{1}+\frac{1}{a}\ge2\sqrt{\frac{a}{1}.\frac{1}{a}}=2.1=2\) (bất đẳng thức Cauchy)
Chứng minh:
\(\left(a-b\right)^2\ge0\left(\forall a,b\right)\)
\(\Leftrightarrow a^2-2ab+b^2\ge0\)
\(\Leftrightarrow a^2+2ab+b^2\ge4ab\)
\(\Leftrightarrow\left(a+b\right)^2\ge4ab\)
\(\Leftrightarrow a+b\ge2\sqrt{ab}\)
(áp dụng vào cái trên)
Dấu "=" xảy ra khi:
\(a=\frac{1}{a}\Leftrightarrow a^2=1\Rightarrow a=1\left(a>0\right)\)
AM-GM :\(\dfrac{1}{a^4+b^2+2ab^2}=\dfrac{1}{a^4+b^2+ab^2+ab^2}\le\dfrac{1}{4\sqrt[4]{a^6b^6}}\)
\(\Rightarrow Q\le\dfrac{1}{2\sqrt[4]{a^6b^6}}\) (1)
AM - GM : \(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{2}{\sqrt{ab}}\Leftrightarrow2\ge\dfrac{2}{\sqrt{ab}}\Leftrightarrow ab\ge1\) (2)
Kết hợp (1) và (2) ta có đpcm
Ta có: \(\dfrac{1}{1+x^2}+\dfrac{1}{1+y^2}\ge\dfrac{2}{1+xy}\)
\(\Leftrightarrow\left(\dfrac{1}{1+x^2}-\dfrac{1}{1+y^2}\right)+\left(\dfrac{1}{1+y^2}-\dfrac{1}{xy}\right)\ge0\)
\(\Leftrightarrow\dfrac{xy-x^2}{\left(1+x^2\right)\left(1+xy\right)}+\dfrac{xy-y^2}{\left(1+y^2\right)\left(1+xy\right)}\ge0\)
\(\Leftrightarrow\dfrac{x\left(y-x\right)}{\left(1+x^2\right)\left(1+xy\right)}+\dfrac{y\left(x-y\right)}{\left(1+y^2\right)\left(1+xy\right)}\ge0\)
\(\Leftrightarrow\dfrac{\left(y-x\right)^2\left(xy-1\right)}{\left(1+x^2\right)\left(1+y^2\right)\left(1+xy\right)}\ge0\)
BĐT cuối đúng vì x.y > 0 => đpcm
\(S=a^2+\dfrac{1}{a^2}\)
\(S=\dfrac{1}{16}a^2+\dfrac{1}{a^2}+\dfrac{15}{16}a^2\)
\(S\ge2\sqrt{\dfrac{1}{16}a^2\cdot\dfrac{1}{a^2}}+\dfrac{15}{16}\cdot2^2\)
\(S\ge2\cdot\dfrac{1}{4}+\dfrac{15}{4}\)
\(S\ge\dfrac{17}{4}\)
Vậy \(MINS=\dfrac{17}{4}\Leftrightarrow a=2\)
\(S=\dfrac{a}{8}+\dfrac{a}{8}+\dfrac{1}{a^2}+\dfrac{3a}{4}\ge3\sqrt[3]{\dfrac{a^2}{8a^2}}+\dfrac{3\cdot2}{4}=\dfrac{3}{4}+\dfrac{3}{2}=\dfrac{9}{4}\)
Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}\dfrac{a}{8}=\dfrac{1}{a^2}\\a=2\end{matrix}\right.\Leftrightarrow a=2\)