Với các số thực dương a,b thay đổi , hãy tìm GTLN của \(S=\left(a+b\right)\left(\frac{1}{\sqrt{a^2-ab+2b^2}}+\frac{1}{\sqrt{b^2-ab+2b^2}}\right)\)
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20 tháng 5 2019
Ta có:\(7\left(\frac{1}{a^2}+...\right)=6\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right)+2015\)
Mà \(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\le\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\)
=> \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\le2015\)=> \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\le\sqrt{6045}\)
\(P=\frac{1}{\sqrt{3\left(2a^2+b^2\right)}}+...\)
Mà \(\left(2+1\right)\left(2a^2+b^2\right)\ge\left(2a+b\right)^2\)(bất dẳng thức buniacoxki)
=> \(P\le\frac{1}{2a+b}+\frac{1}{2b+c}+\frac{1}{2c+a}\)
Lại có \(\frac{1}{2a+b}=\frac{1}{a+a+b}\le\frac{1}{9}\left(\frac{1}{a}+\frac{1}{a}+\frac{1}{b}\right)\)
=> \(P\le\frac{1}{9}\left(\frac{3}{a}+\frac{3}{b}+\frac{3}{c}\right)\le\frac{\sqrt{6045}}{3}\)
Vậy \(MaxP=\frac{\sqrt{6045}}{3}\)khi \(a=b=c=\frac{\sqrt{6045}}{2015}\)
12 tháng 2 2018
p \(\ge\)\(\frac{4}{a^2+b^2+2\left(a+b\right)}\) +\(\sqrt{\left(1+ab\right)^2}\) (bunhia và cosi)
=\(\frac{4}{a^2+b^2+2ab}+1+ab=\frac{4}{\left(a+b\right)^2}+a+b+1\)
do \(a+b=ab\le\frac{\left(a+b\right)^2}{4}\Rightarrow a+b\ge4\)
dạt a+b = t thì t>=4
cần tìm min \(\frac{4}{t^2}+t+1=\frac{4}{t^2}+\frac{t}{16}+\frac{t}{16}+\frac{7t}{8}+1\)
\(\ge3.\sqrt[3]{\frac{4}{t^2}.\frac{t}{16}.\frac{t}{16}}+\frac{7.4}{8}+1=\frac{21}{4}\)
dau = xay ra khi a=b=2
VT
21 tháng 10 2019
1) Áp dụng bunhiacopxki ta được \(\sqrt{\left(2a^2+b^2\right)\left(2a^2+c^2\right)}\ge\sqrt{\left(2a^2+bc\right)^2}=2a^2+bc\), tương tự với các mẫu ta được vế trái \(\le\frac{a^2}{2a^2+bc}+\frac{b^2}{2b^2+ac}+\frac{c^2}{2c^2+ab}\le1< =>\)\(1-\frac{bc}{2a^2+bc}+1-\frac{ac}{2b^2+ac}+1-\frac{ab}{2c^2+ab}\le2< =>\)
\(\frac{bc}{2a^2+bc}+\frac{ac}{2b^2+ac}+\frac{ab}{2c^2+ab}\ge1\)<=> \(\frac{b^2c^2}{2a^2bc+b^2c^2}+\frac{a^2c^2}{2b^2ac+a^2c^2}+\frac{a^2b^2}{2c^2ab+a^2b^2}\ge1\) (1)
áp dụng (x2 +y2 +z2)(m2+n2+p2) \(\ge\left(xm+yn+zp\right)^2\)
(2a2bc +b2c2 + 2b2ac+a2c2 + 2c2ab+a2b2). VT\(\ge\left(bc+ca+ab\right)^2\) <=> (ab+bc+ca)2. VT \(\ge\left(ab+bc+ca\right)^2< =>VT\ge1\) ( vậy (1) đúng)
dấu '=' khi a=b=c
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 \(7\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)=6\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)+2015\)
=> \(a=b=c=\sqrt{\frac{3}{2015}}\)
CM
23 tháng 11 2020
1)
Ta có: \(M=\Sigma_{cyc}\frac{\sqrt{3}\left(a+b+4c\right)}{\sqrt{3\left(a+b\right)\left(a+b+4c\right)}}\ge\Sigma_{cyc}\frac{\sqrt{3}\left(a+b+4c\right)}{\frac{3\left(a+b\right)+\left(a+b+4c\right)}{2}}=\Sigma_{cyc}\frac{\sqrt{3}\left(a+b+4c\right)}{2\left(a+b+c\right)}=3\sqrt{3}\)
Dấu "=" xảy ra khi a=b=c
CM
24 tháng 11 2020
2)
\(\Sigma_{cyc}\sqrt[3]{\left(\frac{2a}{ab+1}\right)^2}=\Sigma_{cyc}\frac{2a}{\sqrt[3]{2a\left(ab+1\right)^2}}\ge\Sigma_{cyc}\frac{2a}{\frac{2a+\left(ab+1\right)+\left(ab+1\right)}{3}}=3\Sigma_{cyc}\frac{a}{ab+a+1}\)
Ta có bổ đề: \(\frac{a}{ab+a+1}+\frac{b}{bc+b+1}+\frac{c}{ca+c+1}=1\left(abc=1\right)\)
\(\Rightarrow\Sigma_{cyc}\sqrt[3]{\left(\frac{2a}{ab+1}\right)^2}\ge3\)
ML
18 tháng 8 2016
\(3\left(2a^2+b^2\right)=\left(1^2+1^2+1^2\right)\left(a^2+a^2+b^2\right)\ge\left(a+a+b\right)^2=\left(2a+b\right)^2\)
\(P\le\frac{1}{2a+b}+\frac{1}{2b+c}+\frac{1}{2c+a}\)
\(\frac{1}{2a+b}=\frac{1}{a+a+b}\le\frac{1}{9}\left(\frac{1}{a}+\frac{1}{a}+\frac{1}{b}\right)=\frac{1}{9}\left(\frac{2}{a}+\frac{1}{b}\right)\)
\(P\le\frac{1}{3}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
Đặt \(\left(\frac{1}{a};\frac{1}{b};\frac{1}{c}\right)=\left(x;y;z\right)\)
\(gt\rightarrow7\left(x^2+y^2+z^2\right)=6\left(xy+yz+zx\right)+2015\)
\(\Leftrightarrow7\left(x+y+z\right)^2=20\left(xy+yz+zx\right)+2015\)
Ta có: \(3\left(xy+yz+zx\right)\le\left(x+y+z\right)^2\)
\(\Rightarrow7\left(x+y+z\right)^2\le\frac{20}{3}\left(x+y+z\right)^2+2015\)
\(\Leftrightarrow\frac{1}{3}\left(x+y+z\right)^2\le2015\)
\(\Leftrightarrow x+y+z\le\sqrt{6045}\)
\(P\le\frac{1}{3}\left(x+y+z\right)\le\frac{\sqrt{6045}}{3}\)
Dấu bằng xảy ra khi \(x=y=z=\frac{\sqrt{6045}}{3}\)hay \(a=b=c=\left(\frac{\sqrt{6045}}{3}\right)^{-1}\)
Đánh sai đề kìa :v \(\frac{1}{\sqrt{b^2-ab+2a^2}}\) mới đúng.
Cho \(a=b\rightarrow S=2\sqrt{2}\). Ta cm đây là gtln của S.
\(S\le\left(a+b\right)\sqrt{2\left(\frac{1}{a^2-ab+2b^2}+\frac{1}{b^2-ab+2a^2}\right)}\le2\sqrt{2}\)
\(\Leftrightarrow\left(5a^2-6ab+5b^2\right)\left(a-b\right)^2\ge0\)(bình phương lên quy đồng là xong)
Đẳng thức xảy ra khi a = b.
Đúng ko đấy ạ, sao em quy đồng lên ra \(20a^2b^2-16\left(a^3b+ab^3\right)+5\left(a^4+b^4\right)\)
Nhưng \(\left(a-b\right)^2\left(5a^2-6ab+5b^2\right)=5\left(a^4+b^4\right)+22a^2b^2-16\left(a^3b+ab^3\right)\)