Cho \(\frac{1}{3}< a< 1\)Tìm GTLN
\(Q=\left(1-a\right)\left(3a-1\right)\)
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PN
31 tháng 8 2020
Xét bđt sau :\(\left(a+b^3\right)\left(m+n\right)\ge\left(\sqrt{am}+\sqrt{b^3n}\right)^2\)(đúng theo bunhia nhé)
Chon \(m=a;n=\frac{1}{b}\)khi đó :
\(\left(a+b^3\right)\left(\frac{1}{a}+b\right)\ge\left(\sqrt{a.a}+\sqrt{b^3.\frac{1}{b}}\right)^2\)
\(< =>\left(a+b^3\right)\left(\frac{1}{a}+b\right)\ge\left(a+b\right)^2\)
\(< =>a+b^3\ge\frac{\left(a+b\right)^2}{\frac{1}{a}+b}=\frac{a\left(a+b\right)^2}{1+ab}\)
Suy ra \(\frac{1}{a+b^3}\le\frac{1+ab}{a\left(a+b\right)^2}\)(*)
Bằng cách chứng minh tương tự ta được :\(\frac{1}{a^3+b}\le\frac{1+ab}{b\left(a+b\right)^2}\)(**)
Từ (*) và (**) suy ra : \(\frac{1}{a+b^3}+\frac{1}{a^3+b}\le\frac{1+ab}{a\left(a+b\right)^2}+\frac{1+ab}{b\left(a+b\right)^2}\)
\(=\frac{1}{\left(a+b\right)^2}\left(\frac{1+ab}{a}+\frac{1+ab}{b}\right)=\frac{1}{\left(a+b\right)^2}\left(\frac{1}{a}+a+\frac{1}{b}+b\right)\)
\(=\frac{\frac{1}{a}+\frac{1}{b}+a+b}{\left(a+b\right)^2}=\frac{\frac{1}{a}+\frac{1}{b}}{\left(a+b\right)^2}+\frac{1}{a+b}=\frac{\frac{a+b}{ab}}{\left(a+b\right)^2}+\frac{1}{a+b}=\frac{1}{ab\left(a+b\right)}+\frac{1}{a+b}\)
Khi đó bài toán trở thành tìm GTLN của biểu thức :
\(A\le S=\left(a+b\right)\left(\frac{1}{ab\left(a+b\right)}+\frac{1}{a+b}\right)-\frac{1}{ab}=\frac{a+b}{ab\left(a+b\right)}+\frac{a+b}{a+b}-\frac{1}{ab}\)
\(=\frac{1}{ab}+1-\frac{1}{ab}=1\)
Vậy \(A_{max}=1\)đạt được khi ...
chuyên KHTN 2017 ?
6 tháng 5 2020
Áp dụng Bunhia cho bộ số (1;1;1) vfa (a;b;c) ta có 3(a2+b2+c2) >= (a+b+c)2
=> 3(2a2+b2) >=(2a+b2); 3(2b2+c2) >= (2b+c)2; 3(2c2+a2) >= (2c+a)2
=> \(P\le\frac{1}{2a+b}+\frac{1}{2b+c}+\frac{1}{2c+a}\)
Ta có \(\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\ge9\Rightarrow\frac{1}{9}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\ge\frac{1}{x+y+z}\)
=> \(P\le\frac{1}{2a+b}+\frac{1}{2b+c}+\frac{1}{2c+b}\le\frac{1}{9}\left[\left(\frac{1}{a}+\frac{1}{a}+\frac{1}{b}\right)+\left(\frac{1}{b}+\frac{1}{b}+\frac{1}{c}\right)+\left(\frac{1}{c}+\frac{1}{c}+\frac{1}{a}\right)\right]\)
=> \(P\le\frac{1}{9}\left(\frac{3}{a}+\frac{3}{b}+\frac{3}{c}\right)=\frac{1}{3}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\left(I\right)\)
Ta có \(10\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)=3\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)+6\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)+2015\)
\(=3\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2+2015\left(II\right)\)
Áp dụng Bunhia cho bộ số (1;1;1) và \(\left(\frac{1}{a};\frac{1}{b};\frac{1}{c}\right)\)
Ta được \(3\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\ge\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2\)\(\Rightarrow\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\ge\frac{1}{3}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2\)
=> \(10\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\ge10\cdot\frac{1}{3}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2\left(III\right)\)
Từ (I)(II)(III) => \(3\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2+2015\ge10\cdot\frac{1}{3}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2\)
\(\Rightarrow\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2\le3\cdot2015\Rightarrow\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2-3\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2\)
\(\Rightarrow\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\le\sqrt{3\cdot2015}\left(IV\right)\)
Từ (I)(IV) => \(P\le\frac{1}{3}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\le\frac{1}{3}\cdot\sqrt{3\cdot2015}=\sqrt{\frac{2015}{3}}\)
Vậy GTNN của P=\(\sqrt{\frac{2015}{3}}\)khi a=b=c và \(7\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)=6\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)+2015\)
=> \(a=b=c=\sqrt{\frac{3}{2015}}\)
9 tháng 8 2020
gt <=> \(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=1\)
Đặt: \(\frac{1}{a}=x;\frac{1}{b}=y;\frac{1}{c}=z\)
=> Thay vào thì \(VT=\frac{\frac{1}{xy}}{\frac{1}{z}\left(1+\frac{1}{xy}\right)}+\frac{1}{\frac{yz}{\frac{1}{x}\left(1+\frac{1}{yz}\right)}}+\frac{1}{\frac{zx}{\frac{1}{y}\left(1+\frac{1}{zx}\right)}}\)
\(VT=\frac{z}{xy+1}+\frac{x}{yz+1}+\frac{y}{zx+1}=\frac{x^2}{xyz+x}+\frac{y^2}{xyz+y}+\frac{z^2}{xyz+z}\ge\frac{\left(x+y+z\right)^2}{x+y+z+3xyz}\)
Có BĐT x, y, z > 0 thì \(\left(x+y+z\right)\left(xy+yz+zx\right)\ge9xyz\)Ta thay \(xy+yz+zx=1\)vào
=> \(x+y+z\ge9xyz=>\frac{x+y+z}{3}\ge3xyz\)
=> Từ đây thì \(VT\ge\frac{\left(x+y+z\right)^2}{x+y+z+\frac{x+y+z}{3}}=\frac{3}{4}\left(x+y+z\right)\ge\frac{3}{4}.\sqrt{3\left(xy+yz+zx\right)}=\frac{3}{4}.\sqrt{3}=\frac{3\sqrt{3}}{4}\)
=> Ta có ĐPCM . "=" xảy ra <=> x=y=z <=> \(a=b=c=\sqrt{3}\)
\(Q=-3a^2+4a-1=-3\left(a^2-2.a.\frac{2}{3}+\frac{4}{9}\right)+\frac{1}{3}=-3\left(a-\frac{2}{3}\right)^2+\frac{1}{3}\le\frac{1}{3}\)
\(\Rightarrow Q_{max}=\frac{1}{3}\) khi \(a=\frac{2}{3}\)