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Đặt: \(a=\frac{1}{x};b=\frac{1}{y};c=\frac{1}{z}\)
\(\Rightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{xyz}\)
\(\Leftrightarrow xy+yz+zx=1\)
Ta có:
\(S=\frac{\frac{1}{x}}{\sqrt{\frac{1}{y}.\frac{1}{z}\left(1+\frac{1}{x^2}\right)}}+\frac{\frac{1}{y}}{\sqrt{\frac{1}{z}.\frac{1}{x}\left(1+\frac{1}{y^2}\right)}}+\frac{\frac{1}{z}}{\sqrt{\frac{1}{x}.\frac{1}{y}\left(1+\frac{1}{z^2}\right)}}\)
\(=\sqrt{\frac{yz}{1+x^2}}+\sqrt{\frac{zx}{1+y^2}}+\sqrt{\frac{xy}{1+z^2}}\)
\(=\sqrt{\frac{yz}{xy+yz+zx+x^2}}+\sqrt{\frac{zx}{xy+yz+zx+y^2}}+\sqrt{\frac{xy}{xy+yz+zx+z^2}}\)
\(=\sqrt{\frac{yz}{\left(x+y\right)\left(x+z\right)}}+\sqrt{\frac{zx}{\left(y+x\right)\left(y+z\right)}}+\sqrt{\frac{xy}{\left(z+x\right)\left(z+y\right)}}\)
\(\le\frac{1}{2}.\left(\frac{y}{x+y}+\frac{z}{x+z}+\frac{z}{y+z}+\frac{x}{x+y}+\frac{x}{z+x}+\frac{y}{z+y}\right)\)
\(=\frac{1}{2}.\left(1+1+1\right)=\frac{3}{2}\)
Dấu = xảy ra khi \(x=y=z=\sqrt{3}\)
Ta có : \(\frac{a}{\sqrt{bc\left(1+a^2\right)}}=\frac{a}{\sqrt{bc+a.abc}}=\frac{a}{\sqrt{bc+a\left(a+b+c\right)}}\)
\(=\frac{a}{\sqrt{bc+a^2+ab+ac}}\)
\(=\frac{a}{\sqrt{\left(a+b\right)\left(a+c\right)}}\)
Áp dụng bđt Cô-si ngược ta có
\(\frac{a}{\sqrt{bc\left(1+a^2\right)}}=\frac{a}{\sqrt{\left(a+b\right)\left(a+c\right)}}\le\frac{1}{2}\left(\frac{a}{a+b}+\frac{a}{a+c}\right)\)
C/m tương tự được \(\frac{b}{\sqrt{ca\left(1+b^2\right)}}\le\frac{1}{2}\left(\frac{b}{a+b}+\frac{b}{b+c}\right)\)
\(\frac{c}{\sqrt{ab\left(1+c^2\right)}}\le\frac{1}{2}\left(\frac{c}{a+c}+\frac{c}{b+c}\right)\)
Cộng 3 vế của các bđt trên lại ta được
\(A\le\frac{1}{2}\left(\frac{a}{a+b}+\frac{b}{a+b}+\frac{a}{a+c}+\frac{c}{a+c}+\frac{b}{b+c}+\frac{c}{b+c}\right)\)
\(=\frac{3}{2}\)
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}a+b+c=abc\\a=b=c\end{cases}}\Leftrightarrow\hept{\begin{cases}3a=a^3\\a=b=c\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}a^3-3a=0\\a=b=c\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}a\left(a^2-3\right)=0\\a=b=c\end{cases}}\)
\(\Leftrightarrow a=b=c=\sqrt{3}\left(a,b,c>0\right)\)
Vậy \(A_{max}=\frac{3}{2}\Leftrightarrow x=y=z=\sqrt{3}\)
gt <=> \(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=1\)
Đặt: \(\frac{1}{a}=x;\frac{1}{b}=y;\frac{1}{c}=z\)
=> Thay vào thì \(VT=\frac{\frac{1}{xy}}{\frac{1}{z}\left(1+\frac{1}{xy}\right)}+\frac{1}{\frac{yz}{\frac{1}{x}\left(1+\frac{1}{yz}\right)}}+\frac{1}{\frac{zx}{\frac{1}{y}\left(1+\frac{1}{zx}\right)}}\)
\(VT=\frac{z}{xy+1}+\frac{x}{yz+1}+\frac{y}{zx+1}=\frac{x^2}{xyz+x}+\frac{y^2}{xyz+y}+\frac{z^2}{xyz+z}\ge\frac{\left(x+y+z\right)^2}{x+y+z+3xyz}\)
Có BĐT x, y, z > 0 thì \(\left(x+y+z\right)\left(xy+yz+zx\right)\ge9xyz\)Ta thay \(xy+yz+zx=1\)vào
=> \(x+y+z\ge9xyz=>\frac{x+y+z}{3}\ge3xyz\)
=> Từ đây thì \(VT\ge\frac{\left(x+y+z\right)^2}{x+y+z+\frac{x+y+z}{3}}=\frac{3}{4}\left(x+y+z\right)\ge\frac{3}{4}.\sqrt{3\left(xy+yz+zx\right)}=\frac{3}{4}.\sqrt{3}=\frac{3\sqrt{3}}{4}\)
=> Ta có ĐPCM . "=" xảy ra <=> x=y=z <=> \(a=b=c=\sqrt{3}\)
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Câu 1 chuyên phan bội châu
câu c hà nội
câu g khoa học tự nhiên
câu b am-gm dựa vào hằng đẳng thử rồi đặt ẩn phụ
câu f đặt \(a=\frac{2m}{n+p};b=\frac{2n}{p+m};c=\frac{2p}{m+n}\)
Gà như mình mấy câu còn lại ko bt nha ! để bạn tth_pro full cho nhé !
Câu c quen thuộc, chém trước:
Ta có BĐT phụ: \(\frac{x^3}{x^3+\left(y+z\right)^3}\ge\frac{x^4}{\left(x^2+y^2+z^2\right)^2}\) \((\ast)\)
Hay là: \(\frac{1}{x^3+\left(y+z\right)^3}\ge\frac{x}{\left(x^2+y^2+z^2\right)^2}\)
Có: \(8(y^2+z^2) \Big[(x^2 +y^2 +z^2)^2 -x\left\{x^3 +(y+z)^3 \right\}\Big]\)
\(= \left( 4\,x{y}^{2}+4\,x{z}^{2}-{y}^{3}-3\,{y}^{2}z-3\,y{z}^{2}-{z}^{3 } \right) ^{2}+ \left( 7\,{y}^{4}+8\,{y}^{3}z+18\,{y}^{2}{z}^{2}+8\,{z }^{3}y+7\,{z}^{4} \right) \left( y-z \right) ^{2} \)
Từ đó BĐT \((\ast)\) là đúng. Do đó: \(\sqrt{\frac{x^3}{x^3+\left(y+z\right)^3}}\ge\frac{x^2}{x^2+y^2+z^2}\)
\(\therefore VT=\sum\sqrt{\frac{x^3}{x^3+\left(y+z\right)^3}}\ge\sum\frac{x^2}{x^2+y^2+z^2}=1\)
Done.
Xét biểu thức \(\frac{1}{a+2}+\frac{1}{b+2}+\frac{1}{c+2}\)
\(=\frac{\left(a+2\right)\left(b+2\right)+\left(b+2\right)\left(c+2\right)+\left(c+2\right)\left(a+2\right)}{\left(a+2\right)\left(b+2\right)\left(c+2\right)}\)
\(=\frac{\left(ab+bc+ca\right)+4\left(a+b+c\right)+12}{abc+2\left(ab+bc+ca\right)+4\left(a+b+c\right)+8}\)
\(=\frac{\left(ab+bc+ca\right)+4\left(a+b+c\right)+12}{\left(abc+ab+bc+ca\right)+\left(ab+bc+ca\right)+4\left(a+b+c\right)+8}\)
\(=\frac{\left(ab+bc+ca\right)+4\left(a+b+c\right)+12}{4+\left(ab+bc+ca\right)+4\left(a+b+c\right)+8}\)(Do \(ab+bc+ca+abc=4\)theo giả thiết)
\(=\frac{\left(ab+bc+ca\right)+4\left(a+b+c\right)+12}{\left(ab+bc+ca\right)+4\left(a+b+c\right)+12}=1\)(***)
Với x,y dương ta có 2 bất đẳng thức phụ sau:
\(2\left(x^2+y^2\right)\ge\left(x+y\right)^2\)(*)
\(\frac{1}{x+y}\le\frac{1}{4}\left(\frac{1}{x}+\frac{1}{y}\right)\)(**)
Áp dụng (*) và (**), ta có:
\(\frac{1}{\sqrt{2\left(a^2+b^2\right)}+4}\le\frac{1}{a+b+4}=\frac{1}{\left(a+2\right)+\left(b+2\right)}\)
\(\le\frac{1}{4}\left(\frac{1}{a+2}+\frac{1}{b+2}\right)\)(1)
Tương tự ta có: \(\frac{1}{\sqrt{2\left(b^2+c^2\right)}+4}\le\frac{1}{4}\left(\frac{1}{b+2}+\frac{1}{c+2}\right)\)(2)
\(\frac{1}{\sqrt{2\left(c^2+a^2\right)}+4}\le\frac{1}{4}\left(\frac{1}{c+2}+\frac{1}{a+2}\right)\)(3)
Cộng từng vế của các bất đẳng thức (1), (2), (3), ta được:
\(P\le\frac{1}{2}\left(\frac{1}{a+2}+\frac{1}{b+2}+\frac{1}{c+2}\right)=\frac{1}{2}\)(theo (***))
Đẳng thức xảy ra khi \(a=b=c\)
a b c la : nhau vay a 2 b 5 c 9
dap an laf a 4 b 6c 14