Bài lớp 8 thật hả? :(

\(\frac{a}{a+b+1}+\frac{b}{b+c+1}+\frac{c}{c+a+1}\le1\)

\(\Leftrightarrow\frac{a}{4-a}+\frac{b}{4-b}+\frac{c}{4-c}\le1\)

\(\Leftrightarrow a\left(4-b\right)\left(4-c\right)+b\left(4-a\right)\left(4-c\right)+c\left(4-a\right)\left(4-b\right)\le\left(4-a\right)\left(4-b\right)\left(4-c\right)\)

\(\Leftrightarrow a^2b+ac^2+b^2c+abc\le4\) (1)

Ta cần chứng minh (1)

Không mất tính tổng quát, giả sử \(a\le c\le b\)

\(\Rightarrow a\left(a-c\right)\left(b-c\right)\le0\)

\(\Leftrightarrow a^2b+ac^2\le a^2c+abc\)

\(\Leftrightarrow a^2b+ac^2+b^2c+abc\le a^2c+abc+b^2c+abc\)

\(\Leftrightarrow a^2b+ac^2+b^2c+abc\le c\left(a+b\right)^2\)

\(\Leftrightarrow a^2b+ac^2+b^2c+abc\le\frac{1}{2}.2c\left(a+b\right)\left(a+b\right)\le\frac{1}{2}.\frac{\left(2c+a+b+a+b\right)^3}{27}\)

\(\Leftrightarrow a^2b+ac^2+b^2c+abc\le\frac{1}{2}.\frac{8.3^3}{27}=4\) (đpcm)

Dấu "=" xảy ra khi \(a=b=c=1\)

Do biểu thức đề bài và BĐT đều mang tính đối xứng, không mất tính tổng quát giả sử \(a\ge b\ge c\)

Đặt \(\left(x;y;z\right)=\left(b+c-a;c+a-b;a+b-c\right)\) \(\Rightarrow\left\{{}\begin{matrix}y>0\\z>0\end{matrix}\right.\)

Ta cần chứng minh \(xyz\le1\)

Nếu \(x\le0\) thì \(xyz\le0\Rightarrow xyz< 1\) BĐT hiển nhiên đúng

Nếu \(x>0\)

\(\Rightarrow\left\{{}\begin{matrix}a=\frac{y+z}{2}\\b=\frac{x+z}{2}\\c=\frac{x+y}{2}\end{matrix}\right.\) \(\Rightarrow x+y+z=\frac{2}{x+y}+\frac{2}{y+z}+\frac{2}{z+x}\)

\(\Rightarrow x+y+z\le\frac{1}{\sqrt{xy}}+\frac{1}{\sqrt{yz}}+\frac{1}{\sqrt{zx}}\)

\(\Leftrightarrow\sqrt{xyz}\left(x+y+z\right)\le\sqrt{x}+\sqrt{y}+\sqrt{z}\)

\(\Leftrightarrow xyz\left(x+y+z\right)^2\le\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)^2\le3\left(x+y+z\right)\)

\(\Leftrightarrow xyz\left(x+y+z\right)\le3\)

\(\Leftrightarrow xyz.3\sqrt[3]{xyz}\le xyz\left(x+y+z\right)\le3\)

\(\Leftrightarrow xyz\sqrt[3]{xyz}\le1\Leftrightarrow xyz\le1\) (đpcm)

Dấu "=" xảy ra khi \(x=y=z=1\) hay \(a=b=c=1\)

Có: \(VT=\frac{\left(a+c\right)\left(b+c\right)}{a+b}+\frac{\left(a+b\right)\left(c+a\right)}{b+c}+\frac{\left(c+b\right)\left(a+b\right)}{a+c}\) (thay a+ b+c=1 vào r phân tích thành nhân tử)

Lại có: Theo Cô si \(\frac{\left(a+c\right)\left(b+c\right)}{a+b}+\frac{\left(a+b\right)\left(c+a\right)}{b+c}\ge2\left(c+a\right)\)

Tương tự với hai BĐT còn lại và cộng theo vế được: \(2VT\ge4\Leftrightarrow VT\ge2^{\left(đpcm\right)}\)

"=" <=> a = b = c = 1/3

Đặt \(P=\frac{ab+c}{a+b}+\frac{bc+a}{b+c}+\frac{ac+b}{a+c}=\frac{ab+c\left(a+b+c\right)}{a+b}+\frac{bc+a\left(a+b+c\right)}{b+c}+\frac{ac+b\left(a+b+c\right)}{a+c}\)

\(=\frac{\left(a+c\right)\left(b+c\right)}{a+b}+\frac{\left(a+b\right)\left(a+c\right)}{b+c}+\frac{\left(a+b\right)\left(b+c\right)}{a+c}\)

Ta có:

\(\frac{\left(a+c\right)\left(b+c\right)}{a+b}+\frac{\left(a+b\right)\left(a+c\right)}{b+c}\ge2\left(a+c\right)\)

\(\frac{\left(a+b\right)\left(a+c\right)}{b+c}+\frac{\left(a+b\right)\left(b+c\right)}{a+c}\ge2\left(a+b\right)\)

\(\frac{\left(a+c\right)\left(b+c\right)}{a+b}+\frac{\left(a+b\right)\left(b+c\right)}{a+c}\ge2\left(b+c\right)\)

Cộng vế với vế

\(2P\ge4\left(a+b+c\right)=4\Rightarrow P\ge2\)

Dấu "=" xảy ra khi \(a=b=c=\frac{1}{3}\)

neu de bai bai 1 la tinh x+y thi mik lam cho

đăng từng này thì ai làm cho 

We have \(P=\frac{x^4+2x^2+2}{x^2+1}\)

\(\Rightarrow P=\frac{x^4+2x^2+1+1}{x^2+1}\)

\(=\frac{\left(x^2+1\right)^2+1}{x^2+1}\)

\(=\left(x^2+1\right)+\frac{1}{x^2+1}\)

\(\ge2\sqrt{\frac{x^2+1}{x^2+1}}=2\)

(Dấu "="\(\Leftrightarrow x=0\))

Vậy \(P_{min}=2\Leftrightarrow x=0\)

Ta có: \(\frac{a+7}{a-7}=\frac{b+4}{b-4}\Rightarrow\left(a+7\right)\left(b-4\right)=\left(a-7\right)\left(b+4\right)\)

\(\Rightarrow ab-4a+7b-28=ab+4a-7b-28\)

\(\Rightarrow-4a-4a+7b+7b=0\Rightarrow-8a+14b=0\)

\(\Rightarrow8a=14b\Rightarrow4a=7b\Rightarrow\frac{a}{7}=\frac{b}{4}\Rightarrow\frac{a}{35}=\frac{b}{20}\left(1\right)\)

Lại có: \(\frac{b+5}{b-5}=\frac{c+6}{c-6}\Rightarrow\left(b+5\right)\left(c-6\right)=\left(b-5\right)\left(c+6\right)\)

\(\Rightarrow bc-6b+5c-30=bc+6b-5c-30\)

\(\Rightarrow6b+6b=5c+5c\) => 12b = 10c

=>\(6b=5c\Rightarrow\frac{b}{5}=\frac{c}{6}\Rightarrow\frac{b}{20}=\frac{c}{24}\left(2\right)\)

Từ \(\left(1\right),\left(2\right)\Rightarrow\frac{a}{35}=\frac{b}{20}=\frac{c}{24}=\frac{a+b+c}{35+20+24}=\frac{79}{79}=1\)

=>a=35,b=20,c=24

cảm ơn bạn

Áp dụng bđt \(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\left(x;y>0\right)\) (tự c/m ha)

\(\frac{7}{a}+\frac{5}{b}+\frac{4}{c}=\left(\frac{4}{a}+\frac{4}{b}\right)+\left(\frac{1}{b}+\frac{1}{c}\right)+\left(\frac{3}{a}+\frac{3}{c}\right)\)

                               \(=4\left(\frac{1}{a}+\frac{1}{b}\right)+\left(\frac{1}{b}+\frac{1}{c}\right)+3\left(\frac{1}{a}+\frac{1}{c}\right)\)

                               \(\ge4.\frac{4}{a+b}+\frac{4}{b+c}+3.\frac{4}{a+c}=4\left(\frac{4}{a+b}+\frac{1}{b+c}+\frac{3}{c+a}\right)\)

Dấu "=" <=> a = b = c

Đặt \(\hept{\begin{cases}x=3a+b+c\\y=3b+a+c\\z=3c+a+b\end{cases}\left(x;y;z>0\right)}\)

\(\Rightarrow x+y+z=5a+5b+5c=5\left(a+b+c\right)\)

Lại có: \(a+b+c=x-2a=y-2b=z-2c\)

\(\Rightarrow x+y+z=5\left(x-2a\right)=5\left(y-2b\right)=5\left(z-2c\right)\)

\(\Rightarrow4x-\left(y+z\right)=4\left(3a+b+c\right)-\left(4b+4c+2a\right)=10a\)

Tương tự ta có:\(4y-\left(x+z\right)=10b;4z-\left(x+y\right)=10c\)

\(\Rightarrow10T=\frac{4x-\left(y+z\right)}{x}+\frac{4y-\left(x+z\right)}{y}+\frac{4z-\left(x+y\right)}{z}\)

\(=12-\frac{y+z}{x}+\frac{x+z}{y}+\frac{x+y}{z}\)

\(=12-\left(\frac{y}{x}+\frac{z}{x}+\frac{x}{y}+\frac{z}{y}+\frac{x}{z}+\frac{y}{z}\right)\)\(\le12-6=6\)(Bđt Cô si)

\(\Rightarrow10T\le6\Rightarrow T\le\frac{6}{10}=\frac{3}{5}\)(Đpcm)

Dấu = khi a=b=c