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cmtt \(\frac{b^2}{a+c}+\frac{a+c}{4}\ge b\)
\(\frac{c^2}{a+b}+\frac{a+b}{4}\ge c\)
\(\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{a+b}+\frac{a+b+b+c+c+a}{4}\ge a+b+c\)
\(A+\frac{1}{2}\ge1\)
\(\frac{a^2}{b+c}+\frac{b+c}{4}\ge2\sqrt{\frac{a^2}{4}}=a\)
cmtt
A+1/2\(\ge1\Rightarrow A\ge\frac{1}{2}\)
A là biểu thức bên trái nha
\(a+b+c=abc\Leftrightarrow\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=1\)
Đặt \(\left(\frac{1}{a};\frac{1}{b};\frac{1}{c}\right)=\left(x;y;z\right)\Rightarrow xy+yz+zx=1\)
\(VT=\frac{x^2yz}{1+yz}+\frac{xy^2z}{1+zx}+\frac{xyz^2}{1+xy}=\frac{x^2yz}{xy+yz+yz+zx}+\frac{xy^2z}{xy+zx+yz+zx}+\frac{xyz^2}{xy+yz+xy+zx}\)
\(VT\le\frac{1}{4}\left(\frac{x^2yz}{xy+yz}+\frac{x^2yz}{yz+zx}+\frac{xy^2z}{xy+zx}+\frac{xy^2z}{yz+zx}+\frac{xyz^2}{xy+yz}+\frac{xyz^2}{xy+zx}\right)\)
\(VT\le\frac{1}{4}\left(\frac{x^2y}{x+y}+\frac{xy^2}{x+y}+\frac{y^2z}{y+z}+\frac{yz^2}{y+z}+\frac{x^2z}{x+z}+\frac{xz^2}{x+z}\right)\)
\(VT\le\frac{1}{4}\left(xy+yz+zx\right)=\frac{1}{4}\)
Dấu "=" xảy ra khi \(a=b=c=\sqrt{3}\)
Cho a,b,c la cac so duong a+b+c=3
Chung minh:\(a^5+b^5+c^5+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge6\)
Áp dụng bđt AM-GM:
\(a^5+\frac{1}{a}\ge2\sqrt{a^5.\frac{1}{a}}=2a^2\)
\(b^5+\frac{1}{b}\ge2\sqrt{b^5.\frac{1}{b}}=2b^2\)
\(c^5+\frac{1}{c}\ge2\sqrt{c^5.\frac{1}{c}}=2c^2\)
\(\Rightarrow VT\ge2\left(a^2+b^2+c^2\right)\ge\frac{2}{3}\left(a+b+c\right)^2=6\)
\("="\Leftrightarrow a=b=c=1\)
chuẩn hóa \(a^2+b^2+c^2=1\)
\(VT\ge\frac{3\sqrt{3}}{2}.\)
chúng ta cần chứng minh:\(\frac{a}{b^2+c^2}\ge\frac{3\sqrt{3}a^2}{2}\Leftrightarrow\frac{a}{1-a^2}\ge\frac{3\sqrt{3}a^2}{2}\)
\(\Leftrightarrow\frac{1}{1-a^2}\ge\frac{3\sqrt{3}a}{2}.\)
\(\Leftrightarrow a\left(1-a^2\right)\le\frac{2}{3\sqrt{3}}.\)
\(\Leftrightarrow a^2\left(1-a^2\right)^2\le\frac{4}{27}.\)
Mà\(\)
\(\Leftrightarrow2a^2\left(1-a^2\right)\left(1-a^2\right)\le\frac{\left(2a^2+1-a^2+1-a^2\right)^3}{27}=\frac{8}{27}.\left(dung\right)\)
Nên\(a^2\left(1-a^2\right)^2\le\frac{4}{27}\left(luondung\right)\)
Tương tự ta có: \(\frac{b}{a^2+c^2}\ge\frac{3\sqrt{3}b^2}{2};\frac{c}{a^2+b^2}\ge\frac{3\sqrt{3}c^2}{2}\)
Cộng lại ta có \(đpcm\)
Dấu bằng xảy ra khi \(a=b=c=\frac{1}{\sqrt{3}}\)
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1\Leftrightarrow ab+bc+ca=abc\)
\(\frac{a^2}{a+bc}=\frac{a^3}{a^2+abc}=\frac{a^3}{a^2+ab+bc+ca}=\frac{a^3}{\left(a+b\right)\left(a+c\right)}\)
Cô si:
\(\frac{a^3}{\left(a+b\right)\left(a+c\right)}+\frac{a+b}{8}+\frac{a+b}{8}\ge3\sqrt[3]{\frac{a^3}{\left(a+b\right)\left(b+c\right)}.\frac{\left(a+b\right)}{8}.\frac{\left(b+c\right)}{8}}=\frac{3a}{4}\)
Tương tự với 2 cục còn lại, công theo vế:
\(VT+\frac{a+b+c}{2}\ge\frac{3}{4}\left(a+b+c\right)\)
\(\Rightarrow VT\ge\frac{a+b+c}{4}\text{ }\left(dpcm\right)\)