1.
\(ab+1\le b\Leftrightarrow a+\dfrac{1}{b}\le1\)
Đặt \(\left(a;\dfrac{1}{b}\right)=\left(x;y\right)\Rightarrow x+y\le1\)
\(P=x+\dfrac{1}{x^2}+y+\dfrac{1}{y^2}=\left(\dfrac{x}{2}+\dfrac{x}{2}+\dfrac{1}{16x^2}\right)+\left(\dfrac{y}{2}+\dfrac{y}{2}+\dfrac{1}{16y^2}\right)+\dfrac{15}{16}\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}\right)\)
\(P\ge3\sqrt[3]{\dfrac{x^2}{64x^2}}+3\sqrt[3]{\dfrac{y^2}{64y^2}}+\dfrac{15}{16}.\dfrac{1}{2}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)^2\)
\(P\ge\dfrac{3}{2}+\dfrac{15}{32}.\left(\dfrac{4}{x+y}\right)^2\ge\dfrac{3}{2}+\dfrac{15}{32}.4^2=9\)
\(P_{min}=9\) khi \(x=y=\dfrac{1}{2}\) hay \(\left(a;b\right)=\left(\dfrac{1}{2};2\right)\)
2.
\(ab+4\le2b\Rightarrow2\ge a+\dfrac{4}{b}\ge2\sqrt{\dfrac{4a}{b}}\Rightarrow\dfrac{a}{b}\le\dfrac{1}{4}\)
\(\Rightarrow\dfrac{b}{a}\ge4\)
Đặt \(\dfrac{b}{a}=x\Rightarrow x\ge4\)
\(P=\dfrac{\dfrac{b}{a}}{1+2\left(\dfrac{b}{a}\right)^2}=\dfrac{x}{2x^2+1}\Rightarrow\dfrac{1}{P}=2x+\dfrac{1}{x}\)
\(\Rightarrow\dfrac{1}{P}=\left(\dfrac{x}{16}+\dfrac{1}{x}\right)+\dfrac{31}{26}x\ge2\sqrt{\dfrac{x}{16x}}+\dfrac{31}{16}.4=\dfrac{33}{4}\)
\(\Rightarrow P\le\dfrac{4}{33}\)
\(P_{max}=\dfrac{4}{33}\) khi \(x=4\Rightarrow\left(a;b\right)=\left(\dfrac{1}{2};2\right)\)
4.
\(\dfrac{3b}{1+b}=1-\dfrac{a}{1+a}=\dfrac{1}{1+a}\)
Lại có:
\(\dfrac{a}{1+a}+\dfrac{2b}{1+b}=1-\dfrac{b}{1+b}\)
\(\Rightarrow\dfrac{1}{1+b}=\dfrac{a}{1+a}+\dfrac{b}{1+b}+\dfrac{b}{1+b}\ge3\sqrt[3]{\dfrac{ab^2}{\left(1+a\right)\left(1+b\right)^2}}\)
\(\Leftrightarrow\dfrac{1}{\left(1+b\right)^3}\ge\dfrac{27ab^2}{\left(1+a\right)\left(1+b\right)^2}\)
\(\Leftrightarrow\dfrac{1}{1+b}\ge\dfrac{27ab^2}{1+a}=27ab^2.\dfrac{3b}{1+b}\)
\(\Leftrightarrow81ab^3\le1\)
\(\Leftrightarrow ab^3\le\dfrac{1}{81}\)
\(P_{max}=\dfrac{1}{81}\) khi \(a=b=\dfrac{1}{3}\)
3.
\(\dfrac{a}{1+a}+\dfrac{2b}{1+b}=1\Leftrightarrow\dfrac{2b}{1+b}=1-\dfrac{1}{1+a}=\dfrac{1}{1+a}\)
Ta có:
\(\dfrac{a}{1+a}+\dfrac{b}{1+b}=1-\dfrac{b}{1+b}\Rightarrow\dfrac{1}{1+b}=\dfrac{a}{1+a}+\dfrac{b}{1+b}\ge2\sqrt{\dfrac{ab}{\left(1+a\right)\left(1+b\right)}}\)
\(\Rightarrow\dfrac{1}{\left(1+b\right)^2}\ge\dfrac{4ab}{\left(1+a\right)\left(1+b\right)}\Rightarrow\dfrac{1}{1+b}\ge\dfrac{4ab}{1+a}=4ab.\dfrac{2b}{1+b}\)
\(\Rightarrow8ab^2\le1\)
\(\Rightarrow ab^2\le\dfrac{1}{8}\)
Dấu "=" xảy ra khi \(a=b=\dfrac{1}{2}\)
