cho a,b > 0 thỏa mãn a + b = 2. Cmr: \(2\left(a^2+b^2\right)-6\left(\frac{a}{b}+\frac{b}{a}\right)+9\left(\frac{1}{a^2}+\frac{1}{b^2}\right)\ge10\)
Cho hai số thực dương a,b sao cho a+b=2.
CMR: \(2\left(a^2+b^2\right)-6\left(\frac{a}{b}+\frac{b}{a}\right)+9\left(\frac{1}{a^2}+\frac{1}{b^2}\right)\ge10\)
Ta có :
\(P=2\left(a^2+b^2\right)-6\left(\frac{a}{b}+\frac{b}{a}\right)+9\left(\frac{1}{a^2}+\frac{1}{b^2}\right)\)
\(=2a^2+2b^2-\frac{6a}{b}+\frac{6b}{a}+\frac{9}{a^2}+\frac{9}{b^2}\)
\(=\left(\frac{3}{a^2}+3b^2\right)+\left(\frac{3}{b^2}+3a^2\right)-\left(a^2+2ab+b^2\right)-6\left(\frac{a}{b}+\frac{b}{a}\right)+6\left(2ab+\frac{1}{a^2}+\frac{1}{b^2}\right)-10ab\)
\(=\left(\frac{3}{a^2}+3b^2\right)+\left(\frac{3}{b^2}+3a^2\right)-4-6\left(\frac{a}{b}+\frac{b}{a}\right)+6\left(2ab+\frac{1}{a^2}+\frac{1}{b^2}\right)-10ab\)
Áp dụng BĐT Cô si cho các số dương ta có :
\(+,\frac{3}{a^2}+3b^2\ge2\sqrt{\frac{3}{a^2}.3b^2}=\frac{6b}{a}\left(1\right)\)
+, \(\frac{3}{b^2}+3a^2\ge2\sqrt{\frac{3}{b^2}.3a^2}=\frac{6a}{b}\left(2\right)\)
\(+,\left(\frac{a}{b}+\frac{b}{a}\right)\ge2\sqrt{\frac{a}{b}.\frac{a}{b}}=2\Leftrightarrow6\left(\frac{a}{b}+\frac{b}{a}\right)=12\left(3\right)\)
+, \(ab+ab+\frac{1}{a^2}+\frac{1}{b^2}\ge\sqrt{ab.ab.\frac{1}{a^2}.\frac{1}{b^2}}=1\Leftrightarrow6\left(ab+ab+\frac{1}{a^2}+\frac{1}{b^2}\right)=6\)
+) \(ab\ge\frac{\left(a+b\right)^2}{4}\Leftrightarrow10ab\ge10\)
Cộng vế với vế ta có :
\(P\ge10\)
Dấu "=" xảy ra \(\Leftrightarrow a=b\)
Bài 1: Cho a,b,c thỏa mãn a+b+c=3
Tìm GTNN \(P=\frac{a^3}{b\left(2c+a\right)}+\frac{b^3}{c\left(2a+b\right)}+\frac{c^3}{a\left(2b+c\right)}\)
Bài 2: Cho a,b>0 thỏa mãn a+b=2
Tìm GTNN \(Q=2\left(a^2+b^2\right)-6\left(\frac{a}{b}+\frac{b}{a}\right)+9\left(\frac{1}{a^2}+\frac{1}{b^2}\right)\)
bài 1
ÁP dụng AM-GM ta có:
\(\frac{a^3}{b\left(2c+a\right)}+\frac{2c+a}{9}+\frac{b}{3}\ge3\sqrt[3]{\frac{a^3.\left(2c+a\right).b}{b\left(2c+a\right).27}}=a.\)
tương tự ta có:\(\frac{b^3}{c\left(2a+b\right)}+\frac{2a+b}{9}+\frac{c}{3}\ge b,\frac{c^3}{a\left(2b+c\right)}+\frac{2b+c}{9}+\frac{a}{3}\ge c\)
công tất cả lại ta có:
\(P+\frac{2a+b}{9}+\frac{2b+c}{9}+\frac{2c+a}{9}+\frac{a+b+c}{3}\ge a+b+c\)
\(P+\frac{2\left(a+b+c\right)}{3}\ge a+b+c\)
Thay \(a+b+c=3\)vào ta được":
\(P+2\ge3\Leftrightarrow P\ge1\)
Vậy Min là \(1\)
dấu \(=\)xảy ra khi \(a=b=c=1\)
Cho a;b;c>0 thỏa mãn abc=1. CMR:
\(\frac{a}{\left(ab+a+1\right)^2}+\frac{b}{\left(bc+b+1\right)^2}+\frac{c}{\left(ac+c+1\right)^2}\ge\frac{1}{a+b+c}\)
Áp dụng BĐT Bunhiacopxki, ta có:
\(\left(a+b+c\right)\left(\frac{a}{\left(ab+a+1\right)^2}+\frac{b}{\left(bc+b+1\right)^2}+\frac{c}{\left(ca+c+1\right)^2}\right)\ge\left(\frac{a}{ab+a+1}+\frac{b}{bc+b+1}+\frac{c}{ca+c+1}\right)^2\)
Mà \(\frac{a}{ab+a+1}+\frac{b}{bc+b+1}+\frac{c}{ca+c+1}=\frac{a}{ab+a+abc}+\frac{b}{bc+b+1}+\frac{bc}{abc+bc+b}=\frac{1}{b+1+bc}+\frac{b}{bc+b+1}+\frac{bc}{1+bc+1}=1\)
\(\Rightarrow\left(\frac{a}{\left(ab+a+1\right)^2}+\frac{b}{\left(bc+b+1\right)^2}+\frac{c}{\left(ca+c+1\right)^2}\right)\left(a+b+c\right)\ge1\)
\(\Rightarrow\frac{a}{\left(ab+b+1\right)^2}+\frac{b}{\left(bc+b+1\right)^2}+\frac{c}{\left(ac+c+1\right)^2}\ge\frac{1}{a+b+c}\)
\(\frac{a}{\left(ab+a+1\right)^2}+\frac{b}{\left(bc+b+1\right)^2}+\frac{c}{\left(ac+c+1\right)^2}\ge\frac{1}{a+b+c}\)
ta có \(\frac{a}{ab+a+1}+\frac{b}{bc+b+1}+\frac{c}{ca+c+1}\)
\(=\frac{1}{bc+b+1}+\frac{b}{bc+b+1}+\frac{bc}{bc+b+1}=1\)
đặt \(H=\frac{a}{\left(ab+a+1\right)^2}+\frac{b}{\left(bc+b+1\right)^2}+\frac{c}{\left(ac+c+1\right)^2}\)
áp dụng bất đẳng thức bunhiacopxki ta có
\(H\left(a+b+c\right)\ge\left(\frac{a}{ab+a+1}+\frac{b}{bc+b+1}+\frac{c}{ac+c+1}\right)^2=1\)
\(\Rightarrow H\ge\frac{1}{a+b+c}\)
hay \(\frac{a}{\left(ab+a+1\right)^2}+\frac{b}{\left(bc+b+1\right)^2}+\frac{c}{\left(ac+c+1\right)^2}\ge\frac{1}{a+b+c}\)
cho \(a,b,c>0\) thỏa mãn \(abc=1\) CMR:\(\frac{1}{\left(2+a\right)\left(2+\frac{1}{b}\right)}+\frac{1}{\left(2+b\right)\left(2+\frac{1}{c}\right)}+\frac{1}{\left(2+c\right)\left(2+\frac{1}{a}\right)}\le\frac{1}{3}\)
Với a,b,c > 0 thỏa mãn abc = 1 . CMR:
\(\frac{1}{a^2\left(b+c\right)}+\frac{1}{b^2\left(c+a\right)}+\frac{1}{c^2\left(a+b\right)}\ge\frac{3}{2}\)
Có: \(VT=\frac{abc}{a^2\left(b+c\right)}+\frac{abc}{b^2\left(c+a\right)}+\frac{abc}{c^2\left(a+b\right)}\)
\(=\frac{bc}{ab+ac}+\frac{ac}{bc+ba}+\frac{ab}{ac+bc}\)
Áp dụng bđt \(\frac{a^2}{x}+\frac{b^2}{y}+\frac{c^2}{z}\ge\frac{\left(a+b+c\right)^2}{x+y+z}\)được
\(VT\ge\frac{\left(\sqrt{ab}+\sqrt{ac}+\sqrt{bc}\right)^2}{2\left(ab+bc+ca\right)}\)
Mà\(\left(\sqrt{ab}+\sqrt{ac}+\sqrt{bc}\right)^2\ge3\left(ab+bc+ca\right)\)(Chuyển vế đưa thành tổng bình phương)
\(\Rightarrow VT\ge...\ge\frac{3\left(ab+bc+ca\right)}{2\left(ab+bc+ca\right)}=\frac{3}{2}\)
Dấu "=" khi a=b=c=1
Bài 1:Cho a,b,c là các số nguyên đôi 1 khác nhau thỏa mãn a+b+c=2019.tính giá trị biểu thức
\(M=\frac{a^3}{\left(a+b\right)\left(a-c\right)}+\frac{b^3}{\left(b-a\right)\left(b-c\right)}+\frac{c^3}{\left(c-a\right)\left(c-b\right)}\)
Bài 2:Cho \(a+b+c=0;P=\frac{a-b}{c}+\frac{b-c}{a}+\frac{c-a}{b};Q=\frac{c}{a-b}+\frac{a}{b-c}+\frac{b}{c-a}\)
\(CMR\) \(P\cdot Q=9\)
Bài 3:Cho 3 số x;y;z đôi 1 khác nhau thỏa mãn x+y+z=0 và \(A=\frac{4xy-z^2}{xy+2z^2};B=\frac{4yz-x^2}{yz+2x^2};C=\frac{4xz-y^2}{xz+2y^2}\)
CMR A.B.C=1
Đặt \(\left(\frac{a-b}{c},\frac{b-c}{a},\frac{c-a}{b}\right)\rightarrow\left(x,y,z\right)\)
Khi đó:\(\left(\frac{c}{a-b},\frac{a}{b-c},\frac{b}{c-a}\right)\rightarrow\left(\frac{1}{x},\frac{1}{y},\frac{1}{z}\right)\)
Ta có:
\(P\cdot Q=\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=3+\frac{y+z}{x}+\frac{z+x}{y}+\frac{x+y}{z}\)
Mặt khác:\(\frac{y+z}{x}=\left(\frac{b-c}{a}+\frac{c-a}{b}\right)\cdot\frac{c}{a-b}=\frac{b^2-bc+ac-a^2}{ab}\cdot\frac{c}{a-b}\)
\(=\frac{c\left(a-b\right)\left(c-a-b\right)}{ab\left(a-b\right)}=\frac{c\left(c-a-b\right)}{ab}=\frac{2c^2}{ab}\left(1\right)\)
Tương tự:\(\frac{x+z}{y}=\frac{2a^2}{bc}\left(2\right)\)
\(=\frac{x+y}{z}=\frac{2b^2}{ac}\left(3\right)\)
Từ ( 1 );( 2 );( 3 ) ta có:
\(P\cdot Q=3+\frac{2c^2}{ab}+\frac{2a^2}{bc}+\frac{2b^2}{ac}=3+\frac{2}{abc}\left(a^3+b^3+c^3\right)\)
Ta có:\(a+b+c=0\)
\(\Rightarrow\left(a+b\right)^3=-c^3\)
\(\Rightarrow a^3+b^3+3ab\left(a+b\right)=-c^3\)
\(\Rightarrow a^3+b^3+c^3=3abc\)
Khi đó:\(P\cdot Q=3+\frac{2}{abc}\cdot3abc=9\)
Bài 1 :Cho a,b,c dương thỏa mãn a+b+c=2
CMR \(\frac{bc}{\sqrt{3a^2+4}}+\frac{ca}{\sqrt{3b^2+4}}+\frac{ab}{\sqrt{3c^2+4}}\ge\frac{\sqrt{3}}{3}\)
Bài 2:Cho a,b,c>0. CMR
\(\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge\frac{8}{9}\left(a+b+c\right)\left(ab+bc+ca\right)\)
\(\left(a+b\right)\left(b+c\right)\left(c+a\right)+abc\)
\(=abc+a^2b+ab^2+a^2c+ac^2+b^2c+bc^2+abc+abc\)
\(=\left(a+b+c\right)\left(ab+bc+ca\right)\)( phân tích nhân tử các kiểu )
\(\Rightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge\left(a+b+c\right)\left(ab+bc+ca\right)-abc\left(1\right)\)
\(a+b+c\ge3\sqrt[3]{abc};ab+bc+ca\ge3\sqrt[3]{a^2b^2c^2}\Rightarrow\left(a+b+c\right)\left(ab+bc+ca\right)\ge9abc\)
\(\Rightarrow-abc\ge\frac{-\left(a+b+c\right)\left(ab+bc+ca\right)}{9}\)
Khi đó:\(\left(a+b+c\right)\left(ab+bc+ca\right)-abc\)
\(\ge\left(a+b+c\right)\left(ab+bc+ca\right)-\frac{\left(a+b+c\right)\left(ab+bc+ca\right)}{9}\)
\(=\frac{8\left(a+b+c\right)\left(ab+bc+ca\right)}{9}\left(2\right)\)
Từ ( 1 ) và ( 2 ) có đpcm
Cho các số thực a, b, c đôi một khác nhau thỏa mãn \(0\le a;b;c\le2\)
CMR: \(\frac{1}{^{\left(a-b\right)^2}}+\frac{1}{\left(b-c\right)^2}+\frac{1}{\left(c-a\right)^2}\ge\frac{9}{4}\)
Vì vai trò bình đẳng của các ẩn \(a,b,c\) là như nhau nên không mất tính tổng quát, ta có thể giả sử:
\(2\ge c>b>a\ge0\) \(\left(\alpha\right)\) (do \(a,b,c\) đôi một khác nhau nên cũng không đồng thời bằng nhau)
Áp dụng bđt \(AM-GM\) cho từng bộ số gồm có các số không âm, ta có:
\(\left(i\right)\) Với \(\frac{1}{\left(a-b\right)^2}>0;\) \(\left[-\left(a-b\right)\right]>0\)\(\frac{1}{\left(a-b\right)^2}+\left[-\left(a-b\right)\right]+\left[-\left(a-b\right)\right]\ge3\sqrt[3]{\frac{1}{\left(a-b\right)^2}.\left[-\left(a-b\right)\right]\left[-\left(a-b\right)\right]}=3\)
\(\Rightarrow\) \(\frac{1}{\left(a-b\right)^2}\ge3-2\left(b-a\right)\) \(\left(1\right)\)
\(\left(ii\right)\) Với \(\frac{1}{\left(b-c\right)^2}>0;\) \(\left[-\left(b-c\right)\right]>0\)
\(\frac{1}{\left(b-c\right)^2}+\left[-\left(b-c\right)\right]+\left[-\left(b-c\right)\right]\ge3\sqrt[3]{\frac{1}{\left(b-c\right)^2}.\left[-\left(b-c\right)\right]\left[-\left(b-c\right)\right]}=3\)
\(\Rightarrow\) \(\frac{1}{\left(b-c\right)^2}\ge3-2\left(c-b\right)\) \(\left(2\right)\)
\(\left(iii\right)\) Với \(\frac{1}{\left(c-a\right)^2}>0;\) \(\frac{c-a}{16}>0\)
\(\frac{1}{\left(c-a\right)^2}+\frac{c-a}{8}+\frac{c-a}{8}\ge3\sqrt[3]{\frac{1}{\left(c-a\right)^2}.\frac{\left(c-a\right)}{8}.\frac{\left(c-a\right)}{8}}=\frac{3}{4}\)
\(\Rightarrow\) \(\frac{1}{\left(c-a\right)^2}\ge\frac{3}{4}-\frac{c-a}{4}\) \(\left(3\right)\)
Cộng từng vế ba bất đẳng thức \(\left(1\right);\) \(\left(2\right)\) và \(\left(3\right)\) , ta được:
\(\frac{1}{\left(a-b\right)^2}+\frac{1}{\left(b-c\right)^2}+\frac{1}{\left(c-a\right)^2}\ge3-2\left(b-a\right)+3-2\left(c-b\right)+\frac{3}{4}-\frac{c-a}{4}\)
nên \(\frac{1}{\left(a-b\right)^2}+\frac{1}{\left(b-c\right)^2}+\frac{1}{\left(c-a\right)^2}\ge\frac{27}{4}-\frac{9\left(c-a\right)}{4}=\frac{27}{4}+\frac{9\left(a-c\right)}{4}\)
Mặt khác, từ \(\left(\alpha\right)\) ta suy ra được: \(\hept{\begin{cases}a\ge0\\2\ge c\end{cases}}\)
nên \(a+2\ge c\) hay nói cách khác \(a-c\ge-2\)
Do đó, \(\frac{1}{\left(a-b\right)^2}+\frac{1}{\left(b-c\right)^2}+\frac{1}{\left(c-a\right)^2}\ge\frac{27}{4}+\frac{9.\left(-2\right)}{4}=\frac{9}{4}\)
Dấu \("="\) xảy ra khi và chỉ khi \(\hept{\begin{cases}a=0\\b=1\\c=2\end{cases}}\) (thỏa mãn \(\left(\alpha\right)\) )
Vì vai trò bình đẳng của các ẩn \(a,b,c\) là như nhau nên không mất tính tổng quát, ta có thể giả sử:
\(2\ge c>b>a\ge0\) \(\left(\alpha\right)\) (do \(a,b,c\) đôi một khác nhau nên cũng không đồng thời bằng nhau)
Áp dụng bđt \(AM-GM\) cho từng bộ số gồm có các số không âm, ta có:
\(\left(i\right)\) Với \(\frac{1}{\left(a-b\right)^2}>0;\) \(\left[-\left(a-b\right)\right]>0\)\(\frac{1}{\left(a-b\right)^2}+\left[-\left(a-b\right)\right]+\left[-\left(a-b\right)\right]\ge3\sqrt[3]{\frac{1}{\left(a-b\right)^2}.\left[-\left(a-b\right)\right]\left[-\left(a-b\right)\right]}=3\)
\(\Rightarrow\) \(\frac{1}{\left(a-b\right)^2}\ge3-2\left(b-a\right)\) \(\left(1\right)\)
\(\left(ii\right)\) Với \(\frac{1}{\left(b-c\right)^2}>0;\) \(\left[-\left(b-c\right)\right]>0\)
\(\frac{1}{\left(b-c\right)^2}+\left[-\left(b-c\right)\right]+\left[-\left(b-c\right)\right]\ge3\sqrt[3]{\frac{1}{\left(b-c\right)^2}.\left[-\left(b-c\right)\right]\left[-\left(b-c\right)\right]}=3\)
\(\Rightarrow\) \(\frac{1}{\left(b-c\right)^2}\ge3-2\left(c-b\right)\) \(\left(2\right)\)
\(\left(iii\right)\) Với \(\frac{1}{\left(c-a\right)^2}>0;\) \(\frac{c-a}{16}>0\)
\(\frac{1}{\left(c-a\right)^2}+\frac{c-a}{8}+\frac{c-a}{8}\ge3\sqrt[3]{\frac{1}{\left(c-a\right)^2}.\frac{\left(c-a\right)}{8}.\frac{\left(c-a\right)}{8}}=\frac{3}{4}\)
\(\Rightarrow\) \(\frac{1}{\left(c-a\right)^2}\ge\frac{3}{4}-\frac{c-a}{4}\) \(\left(3\right)\)
Cộng từng vế ba bất đẳng thức \(\left(1\right);\) \(\left(2\right)\) và \(\left(3\right)\) , ta được:
\(\frac{1}{\left(a-b\right)^2}+\frac{1}{\left(b-c\right)^2}+\frac{1}{\left(c-a\right)^2}\ge3-2\left(b-a\right)+3-2\left(c-b\right)+\frac{3}{4}-\frac{c-a}{4}\)
nên \(\frac{1}{\left(a-b\right)^2}+\frac{1}{\left(b-c\right)^2}+\frac{1}{\left(c-a\right)^2}\ge\frac{27}{4}-\frac{9\left(c-a\right)}{4}=\frac{27}{4}+\frac{9\left(a-c\right)}{4}\)
Mặt khác, từ \(\left(\alpha\right)\) ta suy ra được: \(\hept{\begin{cases}a\ge0\\2\ge c\end{cases}}\)
nên \(a+2\ge c\) hay nói cách khác \(a-c\ge-2\)
Do đó, \(\frac{1}{\left(a-b\right)^2}+\frac{1}{\left(b-c\right)^2}+\frac{1}{\left(c-a\right)^2}\ge\frac{27}{4}+\frac{9.\left(-2\right)}{4}=\frac{9}{4}\)
Dấu \("="\) xảy ra khi và chỉ khi \(a=0;b=1;c=2\) (thỏa mãn \(\left(\alpha\right)\) )
Cho các số dương a, b, c thỏa mãn ab+bc+ca=1.
CMR: \(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\ge3+\sqrt{\frac{\left(a+b\right)\left(a+c\right)}{a^2}}+\sqrt{\frac{\left(b+c\right)\left(b+a\right)}{b^2}}+\sqrt{\frac{\left(c+a\right)\left(c+b\right)}{c^2}}\)