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Ta có: a , b , c > 0 => a , b , c là 3 số thực dương thỏa mãn điều kiện: ab + ac + bc = 0
Áp dụng tính chất tỉ dãy số bằng nhau ta có:
\(\frac{a^4}{b+3c}+\frac{b^4}{c+3a}+\frac{c^4}{a+3b}=\frac{a^4+b^4+c^4}{b+3+c+3a+a+3b}\)
\(\Leftrightarrow\frac{a^4+b^4+c^4}{4a+4b+4c}=\frac{a^4+b^4+c^4}{4\left(a+b+c\right)}=\frac{3}{4}\) (Đúng với đề bài)
\(\RightarrowĐPCM\)
Ps; Không chắc nha! Mình chưa học lớp 9
Gọi VT = T
Đặt \(x=3a+b+c;y=3b+c+a;z=3c+a+b\)
\(\Rightarrow x+y+z=5\left(a+b+c\right)=5\left(x-2a\right)=5\left(y-2b\right)\)
\(=5\left(z-2c\right)\)
\(\Rightarrow4x-\left(y+z\right)=10a;4y-\left(z+x\right)=10b;4z-\left(x+y\right)=10c\)
\(\Rightarrow10T=\frac{4x-\left(y+z\right)}{x}+\frac{4y-\left(z+x\right)}{y}+\frac{4z-\left(x+y\right)}{z}\)
\(=12-\left(\frac{y}{x}+\frac{z}{x}+\frac{x}{y}+\frac{z}{y}+\frac{x}{z}+\frac{y}{z}\right)\le12-6=6\)
\(\Rightarrow T\le\frac{6}{10}=\frac{3}{5}\)
Dấu "=" khi a = b = c
1. BĐT ban đầu
<=> \(\left(\frac{1}{3}-\frac{b}{a+3b}\right)+\left(\frac{1}{3}-\frac{c}{b+3c}\right)+\left(\frac{1}{3}-\frac{a}{c+3a}\right)\ge\frac{1}{4}\)
<=>\(\frac{a}{a+3b}+\frac{b}{b+3c}+\frac{c}{c+3a}\ge\frac{3}{4}\)
<=> \(\frac{a^2}{a^2+3ab}+\frac{b^2}{b^2+3bc}+\frac{c^2}{c^2+3ac}\ge\frac{3}{4}\)
Áp dụng BĐT buniacoxki dang phân thức
=> BĐT cần CM
<=> \(\frac{\left(a+b+c\right)^2}{a^2+b^2+c^2+3\left(ab+bc+ac\right)}\ge\frac{3}{4}\)
<=> \(a^2+b^2+c^2\ge ab+bc+ac\)luôn đúng
=> BĐT được CM
2) \(a+b+c\le ab+bc+ca\le\frac{\left(a+b+c\right)^2}{3}\)\(\Leftrightarrow\)\(\left(a+b+c\right)^2-3\left(a+b+c\right)\ge0\)
\(\Leftrightarrow\)\(\left(a+b+c\right)\left(a+b+c-3\right)\ge0\)\(\Leftrightarrow\)\(a+b+c\ge3\)
ko mất tính tổng quát giả sử \(a\ge b\ge c\)
Có: \(3\le a+b+c\le ab+bc+ca\le3a^2\)\(\Leftrightarrow\)\(3a^2\ge3\)\(\Leftrightarrow\)\(a\ge1\)
=> \(\frac{1}{1+a+b}+\frac{1}{1+b+c}+\frac{1}{1+c+a}\le\frac{3}{1+2a}\le1\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(a=b=c=1\)
b) Ta có:
\(\frac{a}{\sqrt{b^2+3}}+\frac{a}{\sqrt{b^2+3}}+\frac{b^2+3}{8}+\frac{a^2}{2}\)\(\ge\)\(4\sqrt[4]{\frac{a^4}{16}}=2a\)
\(\frac{b}{\sqrt{c^2+3}}+\frac{b}{\sqrt{c^2+3}}+\frac{c^2+3}{8}+\frac{b^2}{2}\ge4\sqrt[4]{\frac{b^4}{16}}=2b\)
\(\frac{c}{\sqrt{a^2+3}}+\frac{c}{\sqrt{a^2+3}}+\frac{a^2+3}{8}+\frac{c^2}{2}\ge4\sqrt[4]{\frac{c^4}{16}}=2c\)
Cộng lại ta đươc:
\(2\left(\frac{a}{\sqrt{b^2+3}}+\frac{b}{\sqrt{c^2+3}}+\frac{c}{\sqrt{a^2+3}}\right)+\)\(\frac{5\left(a^2+b^2+c^2\right)+9}{8}\)\(\ge2\left(a+b+c\right)\)
⇒ \(2\left(\frac{a}{\sqrt{b^2+3}}+\frac{b}{\sqrt{c^2+3}}+\frac{c}{\sqrt{a^2+3}}\right)\ge\)\(6-\frac{5\left(a^2+b^2+c^2\right)+9}{8}\)(1)
Lại có: \(a^2+1\ge2a\); \(b^2+1\ge2b\); \(c^2+1\ge2c\)
Suy ra \(a^2+b^2+c^2\ge2\left(a+b+c\right)-3=3\)
Khi đó (1)⇔ \(2\left(\frac{a}{\sqrt{b^2+3}}+\frac{b}{\sqrt{c^2+3}}+\frac{c}{\sqrt{a^2+3}}\right)\ge\)\(6-\frac{5.3+9}{8}=3\)
⇒ \(\frac{a}{\sqrt{b^2+3}}+\frac{b}{\sqrt{c^2+3}}+\frac{c}{\sqrt{a^2+3}}\ge\frac{3}{2}\)
Dấu "=" xảy ra ⇔ \(a=b=c=1\)
\(\left(a^2+3b^2\right)\left(1+3\right)\ge\left(a+3b\right)^2\Rightarrow\sqrt{a^2+3b^2}\ge\frac{a+3b}{2}\)
\(\Rightarrow P=\sum\frac{ab}{\sqrt{a^2+3b^2}}\le2\sum\frac{ab}{a+3b}=2\sum\frac{ab}{a+b+b+b}\)
\(\Rightarrow P\le\frac{1}{8}\sum ab\left(\frac{1}{a}+\frac{3}{b}\right)=\frac{1}{8}\sum\left(3a+b\right)=\frac{1}{2}\left(a+b+c\right)=\frac{3}{2}\)
"=" \(\Leftrightarrow a=b=c=1\)
Áp dụng bất đẳng thức Cauchy-Schwartz ta có
\(\frac{ab}{a+3b+2c}=\frac{ab}{\left(a+c\right)+\left(b+c\right)+2b}\le\frac{ab}{9}\left(\frac{1}{a+c}+\frac{1}{b+c}+\frac{1}{2b}\right).\)
Tương tự ta có 2 bất đẳng thức khác nữa
\(\frac{bc}{b+3c+2a}=\frac{bc}{\left(b+a\right)+\left(a+c\right)+2c}\le\frac{bc}{9}\left(\frac{1}{b+a}+\frac{1}{a+c}+\frac{1}{2c}\right).\)
\(\frac{ac}{c+3a+2b}=\frac{ac}{\left(a+b\right)+\left(b+a\right)+2a}\le\frac{ac}{9}\left(\frac{1}{c+b}+\frac{1}{b+a}+\frac{1}{2a}\right).\)
Cộng ba bất đẳng thức lại cho ta \(\frac{ab}{a+3b+2c}+\frac{bc}{b+3c+2a}+\frac{ca}{c+3a+2b}\le\)
\(\le\frac{ab}{9}\left(\frac{1}{a+c}+\frac{1}{b+c}+\frac{1}{2b}\right)+\frac{bc}{9}\left(\frac{1}{b+a}+\frac{1}{a+c}+\frac{1}{2c}\right)+\frac{ac}{9}\left(\frac{1}{c+b}+\frac{1}{b+a}+\frac{1}{2a}\right)\)
\(=\frac{1}{9}\left(\frac{ab}{a+c}+\frac{bc}{a+c}\right)+\frac{1}{9}\left(\frac{ab}{b+c}+\frac{ac}{b+c}\right)+\frac{1}{9}\left(\frac{bc}{a+b}+\frac{ac}{a+b}\right)+\frac{a}{18}+\frac{b}{18}+\frac{c}{18}\)
\(=\frac{a+b+c}{6}.\) (ĐPCM)
Akai HarumaNguyễn Thành Trương
<= 3/4 nha ko phải a+b+c