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Ta có: \(P=\left(a^2+\frac{1}{16a^2}\right)+\left(b^2+\frac{1}{16b^2}\right)+\frac{15}{16}\left(\frac{1}{a^2}+\frac{1}{b^2}\right)\)
sử dụng bđt cô-si có: \(a^2+\frac{1}{16a^2}\ge\frac{1}{2};b^2+\frac{1}{16b^2}\ge\frac{1}{2};\frac{1}{a^2}+\frac{1}{b^2}\ge\frac{2}{ab}=\frac{4}{2ab}\)
Lại có: \(\frac{1}{a^2}+\frac{1}{b^2}\ge\frac{4}{a^2+b^2}\)
\(\Rightarrow2\left(\frac{1}{a^2}+\frac{1}{b^2}\right)\ge4\left(\frac{1}{a^2+b^2}+\frac{1}{2ab}\right)\ge4\frac{4}{a^2+b^2+2ab}=\frac{16}{\left(a+b\right)^2}=16\)
\(\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}\ge8\)
\(\Rightarrow P\ge\frac{1}{2}+\frac{1}{2}+\frac{15}{2}=\frac{17}{2}\)
Dấu '=' xảy ra <=> \(\hept{\begin{cases}a=b\\a+b=1\end{cases}\Leftrightarrow a=b=\frac{1}{2}}\)
Áp dụng bđt cosi cho 3 số dương a,b,c>0
\(a+b+c\ge3\sqrt[3]{abc}\)
\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge3\sqrt[3]{\dfrac{1}{a}.\dfrac{1}{b}.\dfrac{1}{c}}\)
Suy ra\(\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge3\sqrt[3]{abc}.3\sqrt[3]{\dfrac{1}{a}.\dfrac{1}{b}.\dfrac{1}{c}}=9\sqrt[3]{\dfrac{abc}{abc}}=9\)
Vậy \(\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge9\)
Làm lại :v
\(\dfrac{a}{1+b}+\dfrac{b}{1+a}+\dfrac{1}{a+b}\)
\(\ge\dfrac{a}{a+2b}+\dfrac{b}{2a+b}+\dfrac{1}{a+b}\)
\(=\dfrac{a^2}{a^2+2ab}+\dfrac{b^2}{2ab+b^2}+\dfrac{1}{a+b}\)
\(\ge\dfrac{\left(a+b\right)^2}{\left(a+b\right)^2+2ab}+\dfrac{1}{a+b}\ge\dfrac{\left(a+b\right)^2}{\left(a+b\right)^2+\dfrac{\left(a+b\right)^2}{2}}+\dfrac{1}{a+b}\)
\(=\dfrac{\left(a+b\right)^2}{\dfrac{3}{2}\left(a+b\right)^2}+\dfrac{1}{a+b}=\dfrac{2}{3}+\dfrac{1}{a+b}\ge\dfrac{2}{3}+1=\dfrac{5}{3}\)
\("="\Leftrightarrow a=b=\dfrac{1}{2}\)
Thật ra bài này t đã làm rồi,mà méo rảnh đi mò link,bạn rảnh thì có thể tìm nhé
\(A=a^2+b^2+\dfrac{1}{a^2}+\dfrac{1}{b^2}\)
\(A=a^2+\dfrac{1}{16a^2}+b^2+\dfrac{1}{16b^2}+\dfrac{15}{16}\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}\right)\)
\(A\ge2\sqrt{a^2\cdot\dfrac{1}{16a^2}}+2\sqrt{b^2\cdot\dfrac{1}{16b^2}}+\dfrac{15}{16}\cdot2\cdot\sqrt{\dfrac{1}{a^2b^2}}\)
\(A\ge1+\dfrac{15}{8ab}\ge1+\dfrac{15}{2\left(a+b\right)^2}\ge\dfrac{17}{2}\)
"="<=>x=y=0,5
\(x=\dfrac{1}{2}\cdot\left(\dfrac{a}{\sqrt{ab}}+\dfrac{b}{\sqrt{ab}}\right)=\dfrac{a+b}{2\sqrt{ab}}\)
\(2\sqrt{x^2}-1=2\cdot\dfrac{a+b}{2\sqrt{ab}}-1=\dfrac{a+b-\sqrt{ab}}{\sqrt{ab}}\)
\(x-\sqrt{x^2-1}=\dfrac{a+b}{2\sqrt{ab}}-\sqrt{\dfrac{a^2+2ab+b^2}{4ab}-1}\)
\(=\dfrac{a+b}{2\sqrt{ab}}-\dfrac{a-b}{2\sqrt{ab}}=\dfrac{2b}{2\sqrt{ab}}=\dfrac{\sqrt{b}}{\sqrt{a}}\)
\(G=\dfrac{a+b-\sqrt{ab}}{\sqrt{ab}}:\dfrac{\sqrt{b}}{\sqrt{a}}=\dfrac{a+b-\sqrt{ab}}{\sqrt{ab}}\cdot\dfrac{\sqrt{a}}{\sqrt{b}}\)
\(=\dfrac{a+b-\sqrt{ab}}{b}\)