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 \(\dfrac{1}{1+a^2}+\dfrac{1}{1+b^2}\ge\dfrac{2}{1+ab}\left(đpcm\right)\)

c/ Ta có:\(6a-5b=1\)

\(\Rightarrow5b=6a-1\)

Theo đề thì: \(A=4a^2+\left(6a-1\right)^2=40a^2-12a+1\)

\(=\left(\left(2\sqrt{10}a\right)^2-\frac{2.2.\sqrt{10}.3a}{\sqrt{10}}+\frac{9}{10}\right)+\frac{1}{10}\)

\(=\left(2\sqrt{10}a-\frac{3}{\sqrt{10}}\right)^2+\frac{1}{10}\ge\frac{1}{10}\)

còn câu a,b nữa a ơi :((

\(A=x+\frac{1}{x^2}=\frac{x}{8}+\frac{x}{8}+\frac{1}{x^2}+\frac{3x}{4}\ge3\sqrt[3]{\frac{x}{8}.\frac{x}{8}.\frac{1}{x^2}}+\frac{3.2}{4}=\frac{3}{4}+\frac{6}{4}=\frac{9}{4}\) ( áp dụng cô- si cho 3 số không âm )

Dấu "=" xảy ra <=> x = 2

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:

\(B< \dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{\left(2n-1\right)\left(2n+1\right)}\)

\(B< \dfrac{1}{2}\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{\left(2n-1\right)\left(2n+1\right)}\right)\)

\(B< \dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\right)\)

\(B< \dfrac{1}{2}\left(1-\dfrac{1}{2n+1}\right)\)

\(B< \dfrac{1}{2}\left(\dfrac{2n+1}{2n+1}-\dfrac{1}{2n+1}\right)\)

\(B< \dfrac{1}{2}.\dfrac{2n}{2n+1}\)

\(B< \dfrac{2n}{4n+2}\)

\(B< \dfrac{2n}{2\left(2n+1\right)}\)

\(B< \dfrac{n}{2n+1}\)

Câu a :

Theo BĐT cauchy schwar ta có :

\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge\dfrac{\left(1+1+1\right)^2}{x+y+z}=\dfrac{9}{x+y+z}\)

\(\Rightarrow\left(x+y+z\right)\left(\dfrac{9}{x+y+z}\right)\ge9\)

Câu b : Sửa lại đề nha :

Theo BĐT cauchy schwar ta có :

\(\dfrac{1}{a^2+2bc}+\dfrac{1}{b^2+2ca}+\dfrac{1}{c^2+2ab}\ge\dfrac{\left(1+1+1\right)^2}{a^2+b^2+c^2+2\left(ab+bc+ca\right)}=\dfrac{9}{\left(a+b+c\right)^2}\)

Vì \(a+b+c\le\Rightarrow\left(a+b+c\right)^2\le1\)

\(\Rightarrow\) \(\dfrac{9}{\left(a+b+c\right)^2}\ge9\)

Mơn 😊