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Do a,b,c dương nên AD BĐT Cauchy:
\(\frac{1}{1+ab}+\frac{1}{1+bc}+\frac{1}{1+ac}\ge\frac{9}{3+ab+bc+ca}\)ca (1)
a2+b2+c2\(\ge\)ab+bc+ca\(\Rightarrow3+a^2+b^2+c^2\ge3+ab+bc+ca\)
\(\Rightarrow\frac{9}{6}\le\frac{9}{3+ab+bc+ca}\left(a^2+b^2+c^2=3\right)\) (2)
\(\left(1\right),\left(2\right)\Rightarrow P\ge\frac{3}{2}\)
\(\text{Dấu = khi a=b=c=1}\)
Cho a , b , c là các số thực dương tùy ý . Chứng minh rằng :
Bài làm:
Ta có: \(\frac{a^2+b^2}{a+b}+\frac{b^2+c^2}{b+c}+\frac{c^2+a^2}{c+a}\)
\(=\frac{a^2}{a+b}+\frac{b^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{b+c}+\frac{c^2}{c+a}+\frac{a^2}{c+a}\)
\(=\left(\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+a}\right)+\left(\frac{a^2}{c+a}+\frac{b^2}{a+b}+\frac{c^2}{b+c}\right)\)
\(\ge\frac{\left(a+b+c\right)^2}{2\left(a+b+c\right)}+\frac{\left(a+b+c\right)^2}{2\left(a+b+c\right)}\)(Cauchy Schwars)
\(=\frac{\left(a+b+c\right)^2}{a+b+c}\ge\frac{3\left(ab+bc+ca\right)}{a+b+c}\)
Dấu "=" xảy ra khi: \(a=b=c\)
Áp dụng bất đẳng thức Bunhiacopxki ta được
\(\frac{a^2+b^2}{a+b}+\frac{b^2+c^2}{b+c}+\frac{c^2+a^2}{c+a}\ge\frac{\left(a+b\right)^2}{2\left(a+b\right)}+\frac{\left(b+c\right)^2}{2\left(b+c\right)}+\frac{\left(c+a\right)^2}{2\left(c+a\right)}\)
\(\ge\frac{\left(2a+2b+2c\right)^2}{4\left(a+b+c\right)}\ge\frac{12\left(ab+bc+ca\right)}{4\left(a+b+c\right)}=\frac{3\left(ab+bc+ca\right)}{a+b+c}\)( rút gọn 12/4)
Bất đẳng thức được chứng minh
Đẳng thức xảy ra khi và chỉ khi \(a=b=c\)
\(3-P=1-\frac{x}{x+1}+1-\frac{y}{y+1}+1-\frac{z}{z+1}\)
\(=\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\ge\frac{9}{x+y+z+3}=\frac{9}{1+3}=\frac{9}{4}\)
\(\Rightarrow P\le\frac{3}{4}\)
Dấu "=" xảy ra tại \(x=y=z=\frac{1}{3}\)
\(Từ GT, ta có: \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\ge6\) Áp dụng bđt AM - GM, ta lại có: \(\frac{1}{a^2}+1\ge\frac{2}{a};\frac{1}{b^2}+1\ge\frac{2}{b};\frac{1}{c^2}+1\ge\frac{2}{c}\) \(\frac{1}{a^2}+\frac{1}{b^2}\ge\frac{2}{ab};\frac{1}{b^2}+\frac{1}{c^2}\ge\frac{2}{bc};\frac{1}{c^2}+\frac{1}{a^2}\ge\frac{2}{ac}\) Cộng theo vế ta có: \(3\left(\text{∑}\frac{1}{a^2}\right)+3\ge2\left(\text{∑}\frac{1}{a}+\text{∑}\frac{1}{ab}\right)\Leftrightarrow\text{∑}\frac{1} {a^2}\ge3\left(đ\text{pcm}\right)\) \(\text{Dau }"="\Leftrightarrow a=b=c=1\)
Từ GT, ta có: \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\ge6\)
Áp dụng bđt AM - GM, ta lại có:
\(\frac{1}{a^2}+1\ge\frac{2}{a};\frac{1}{b^2}+1\ge\frac{2}{b};\frac{1}{c^2}+1\ge\frac{2}{c}\)
\(\frac{1}{a^2}+\frac{1}{b^2}\ge\frac{2}{ab};\frac{1}{b^2}+\frac{1}{c^2}\ge\frac{2}{bc};\frac{1}{c^2}+\frac{1}{a^2}\ge\frac{2}{ac}\)
Cộng theo vế ta có:
\(3\left(\text{∑}\frac{1}{a^2}\right)+3\ge2\left(\text{∑}\frac{1}{a}+\text{∑}\frac{1}{ab}\right)\Leftrightarrow\text{∑}\frac{1}{a^2}\ge3\left(đ\text{pcm}\right)\)
\(\text{Dau }"="\Leftrightarrow a=b=c=1\)
Bài 2:b) \(9=\left(\frac{1}{a^3}+1+1\right)+\left(\frac{1}{b^3}+1+1\right)+\left(\frac{1}{c^3}+1+1\right)\)
\(\ge3\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\therefore\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\le3\)
Ta sẽ chứng minh \(P\le\frac{1}{48}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2\)
Ai có cách hay?
1/Đặt a=1/x,b=1/y,c=1/z ->x+y+z=1.
2a) \(VT=\frac{\left(\frac{1}{a^3}+\frac{1}{b^3}\right)\left(\frac{1}{a}+\frac{1}{b}\right)}{\frac{1}{a}+\frac{1}{b}}\ge\frac{\left(\frac{1}{a^2}+\frac{1}{b^2}\right)^2}{\frac{1}{a}+\frac{1}{b}}\)
\(=\frac{\left[\frac{\left(a^2+b^2\right)^2}{a^4b^4}\right]}{\frac{a+b}{ab}}=\frac{\left(a^2+b^2\right)^2}{a^3b^3\left(a+b\right)}\ge\frac{\left(a+b\right)^3}{4\left(ab\right)^3}\)
\(\ge\frac{\left(a+b\right)^3}{4\left[\frac{\left(a+b\right)^2}{4}\right]^3}=\frac{16}{\left(a+b\right)^3}\)
kết bạn với mình
\(P=\frac{a^3+b^3+c^3}{2abc}+\frac{a^2c+b^2c}{c^3+abc}+\frac{b^2a+c^2a}{a^3+abc}+\frac{c^2b+a^2b}{b^3+abc}\)
\(\ge\frac{a^3}{2abc}+\frac{b^3}{2abc}+\frac{c^3}{2abc}+\frac{2abc}{c^3+abc}+\frac{2abc}{a^3+abc}+\frac{2abc}{b^3+abc}\)
\(=\left(\frac{a^3}{2abc}+\frac{2abc}{a^3+abc}\right)+\left(\frac{b^3}{2abc}+\frac{2abc}{b^3+abc}\right)+\left(\frac{c^3}{2abc}+\frac{2abc}{c^3+abc}\right)\)
Xét: \(\frac{a^3}{2abc}+\frac{2abc}{a^3+abc}=\frac{a^3}{2abc}+\frac{1}{2}+\frac{1}{\frac{a^3}{2abc}+\frac{1}{2}}-\frac{1}{2}\ge2\sqrt{\left(\frac{a^3}{2abc}+\frac{1}{2}\right).\frac{1}{\frac{a^3}{2abc}+\frac{1}{2}}}-\frac{1}{2}=\frac{3}{2}\)
Tương tự với 2 cặp còn lại
Vậy ta có: \(P\ge\frac{3}{2}+\frac{3}{2}+\frac{3}{2}=\frac{9}{2}\)
"=" xảy ra <=> a=b=c