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3.
\(5a^2+2ab+2b^2=\left(a^2-2ab+b^2\right)+\left(4a^2+4ab+b^2\right)\)
\(=\left(a-b\right)^2+\left(2a+b\right)^2\ge\left(2a+b\right)^2\)
\(\Rightarrow\sqrt{5a^2+2ab+2b^2}\ge2a+b\)
\(\Rightarrow\frac{1}{\sqrt{5a^2+2ab+2b^2}}\le\frac{1}{2a+b}\)
Tương tự \(\frac{1}{\sqrt{5b^2+2bc+2c^2}}\le\frac{1}{2b+c};\frac{1}{\sqrt{5c^2+2ca+2a^2}}\le\frac{1}{2c+a}\)
\(\Rightarrow P\le\frac{1}{2a+b}+\frac{1}{2b+c}+\frac{1}{2c+a}\)
\(\le\frac{1}{9}\left(\frac{1}{a}+\frac{1}{a}+\frac{1}{b}+\frac{1}{b}+\frac{1}{b}+\frac{1}{c}+\frac{1}{c}+\frac{1}{c}+\frac{1}{a}\right)\)
\(=\frac{1}{3}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\le\frac{1}{3}.\sqrt{3\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)}=\frac{\sqrt{3}}{3}\)
\(\Rightarrow MaxP=\frac{\sqrt{3}}{3}\Leftrightarrow a=b=c=\sqrt{3}\)
Đặt \(\left(\sqrt{a};\sqrt{b};\sqrt{c}\right)\rightarrow\left(x;y;z\right)\)\(\Rightarrow\)\(x^2+y^2+z^2=4\)
\(P=\frac{x^3}{x+3y}+\frac{y^3}{y+3z}+\frac{z^3}{z+3x}=\frac{x^4}{x^2+3xy}+\frac{y^4}{y^2+3yz}+\frac{z^4}{z^2+3zx}\)
\(\ge\frac{\left(x^2+y^2+z^2\right)^2}{x^2+y^2+z^2+3\left(xy+yz+zx\right)}\ge\frac{\left(x^2+y^2+z^2\right)^2}{x^2+y^2+z^2+3\left(x^2+y^2+z^2\right)}=\frac{4^2}{4+3.4}=1\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(a=b=c=\frac{2}{\sqrt{3}}\)
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 <=> \(a=b=c=\sqrt{3}\)
Bài toán số 41 có 2 cách làm, tôi làm cách thứ 2
Đặt \(Q=\sqrt{\frac{x}{y+z}}+\sqrt{\frac{y}{x+z}}+\sqrt{\frac{z}{x+y}}\)\(\Rightarrow Q^2=\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}+2\left(\sqrt{\frac{xy}{\left(y+z\right)\left(x+z\right)}}+\sqrt{\frac{yz}{\left(x+z\right)\left(y+z\right)}}+\sqrt{\frac{xz}{\left(x+y\right)\left(y+z\right)}}\right)\)ta thấy rằng \(\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}=\frac{1}{4}\left(\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}\right)\left(xy+yz+zx\right)\)
\(=\frac{x^2+y^2+z^2}{4}+\frac{xyz}{4}\left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\right)\ge\frac{x^2+y^2+z^2}{4}\)
Áp dụng bất đẳng thức AM-GM ta có \(\sqrt{\frac{yx}{\left(z+x\right)\left(x+y\right)}}\ge\frac{2yx}{2\sqrt{\left(xy+yz\right)\left(yz+yx\right)}}\ge\frac{2xy}{2xy+yz+xz}\ge\frac{2xy}{2\left(xy+yz+zx\right)}=\frac{xy}{xy+yz+zx}\)
Tương tự ta có \(\hept{\begin{cases}\sqrt{\frac{yz}{\left(z+x\right)\left(z+y\right)}}\ge\frac{yz}{xy+yz+zx}\\\sqrt{\frac{xz}{\left(x+y\right)\left(y+z\right)}}\ge\frac{xz}{xy+yz+zx}\end{cases}}\)
\(\Rightarrow\sqrt{\frac{xy}{\left(y+z\right)\left(z+x\right)}}+\sqrt{\frac{yz}{\left(z+x\right)\left(x+y\right)}}+\sqrt{\frac{zx}{\left(x+y\right)\left(y+z\right)}}\ge1\)nên \(Q\ge\sqrt{\frac{x^2+y^2+z^2}{4}+2}\)
\(\Rightarrow Q\ge\sqrt{\frac{x^2+y^2+z^2}{2}+4}+\frac{4}{\sqrt{x^2+y^2+z^2}}\)
Đặt \(t=\sqrt{x^2+y^2+z^2}\Rightarrow t\ge\sqrt{xy+yz+zx}=2\)
Xét hàm số g(t)=\(\sqrt{\frac{t^2}{2}+4}+\frac{4}{t}\left(t\ge2\right)\)khi đó ta có
\(g'\left(t\right)=\frac{t}{2\sqrt{\frac{t^2}{2}+4}}-\frac{4}{t^2};g'\left(t\right)=0\Leftrightarrow t^6-32t^2-256=0\Leftrightarrow t=2\sqrt{2}\)
Lập bảng biến thiên ta có min[2;\(+\infty\)) \(g\left(t\right)=g\left(2\sqrt{2}\right)=3\sqrt{2}\)
Hay minS=\(3\sqrt{2}\)<=> a=c=1; b=2
Đặt a=xc; b=cy (x;y >=1)
- Thay x=1 vào giả thiết ta có \(\sqrt{b-c}=\sqrt{b}\Rightarrow c=0\) (không thỏa mãn vì c>0)
- Thay y=1 vào giả thiết ta có \(\sqrt{a-c}=\sqrt{a}\Rightarrow c=0\)( không thỏa mãn vì c>0)
- Xét x,y>1 thay vào giả thiết ta có
\(\sqrt{x-1}+\sqrt{y-1}=\sqrt{xy}\Leftrightarrow x+y-2+2\sqrt{\left(x-1\right)\left(y-1\right)}=xy\)
\(\Leftrightarrow xy-x-y+1-2\sqrt{\left(x-1\right)\left(y-1\right)}+1=0\)
\(\Leftrightarrow\left(\sqrt{\left(x-1\right)\left(y-1\right)}-1\right)^2=0\)
\(\Leftrightarrow\sqrt{\left(x-1\right)\left(y-1\right)}=1\Leftrightarrow xy=x+y\ge2\sqrt{xy}\Rightarrow xy\ge4\)
Biểu thức P được viết lại như sau
\(P=\frac{x}{y+1}+\frac{y}{x+1}+\frac{1}{x+y}+\frac{1}{x^2+y^2}=\frac{x^2}{xy+x}+\frac{y^2}{xy+y}+\frac{1}{x^2+y^2}+\frac{1}{\left(x+y\right)^2-2xy}\)
\(P\ge\frac{\left(x+y\right)^2}{2xy+x+y}+\frac{1}{x+y}+\frac{1}{\left(x+y\right)^2-2xy}=\frac{xy}{3}+\frac{1}{xy}+\frac{1}{x^2y^2-2xy}=\frac{x^3y^3-2x^2y^2+3xy-3}{3\left(x^2y^2-2xy\right)}\)
Đặt t=xy với t>=4
Xét hàm số \(f\left(t\right)=\frac{t^3-2t^2+3t-3}{t^2-2t}\left(t\ge4\right)\)
Ta có \(f'\left(t\right)=\frac{t^4-4t^3+t^2+6t-6}{\left(t^2-2t\right)^2}=\frac{t^3\left(t-4\right)+6\left(t-4\right)+18}{\left(t^2-2t\right)^2}>0\forall t\ge4\)
Lập bảng biến thiên ta có \(minf\left(t\right)=f\left(4\right)=\frac{41}{8}\)
Vậy \(minP=\frac{41}{24}\)khi x=y=z=2 hay a=b=2c
\(\sqrt{a^2+\frac{1}{b^2}}+\sqrt{b^2+\frac{1}{c^2}}+\sqrt{c^2+\frac{1}{a^2}}\)
\(\ge\sqrt{\left(a+b+c\right)^2+\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2}\)
\(\ge\sqrt{\left(a+b+c\right)^2+\frac{81}{\left(a+b+c\right)^2}}\)
\(\ge\sqrt{\left(a+b+c\right)^2+\frac{81}{16\left(a+b+c\right)^2}+\frac{1215}{16\left(a+b+c\right)^2}}\)
\(\ge\sqrt{\frac{2.9}{4}+\frac{1215.4}{16.9}}=\frac{3\sqrt{17}}{2}\)
√a2+1b2 +√b2+1c2 +√c2+1a2
≥√(a+b+c)2+(1a +1b +1c )2
≥√(a+b+c)2+81(a+b+c)2
≥√(a+b+c)2+8116(a+b+c)2 +121516(a+b+c)2
≥√2.94 +1215.416.9 =3√172
\(S\ge3\sqrt[6]{\frac{a^2b^2+1}{ab}.\frac{b^2c^2+1}{bc}.\frac{c^2a^2+1}{ca}}\)
Đặt \(\left\{{}\begin{matrix}2\sqrt{a}-3=x>0\\2\sqrt{b}-3=y>0\\2\sqrt{c}-3=z>0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=\frac{\left(x+3\right)^2}{4}\\b=\frac{\left(y+3\right)^2}{4}\\c=\frac{\left(z+3\right)^2}{4}\end{matrix}\right.\)
\(A=\frac{\left(x+3\right)^2}{4y}+\frac{\left(y+3\right)^2}{4z}+\frac{\left(z+3\right)^2}{4x}\)
\(Q\ge\frac{\left(x+y+z+9\right)^2}{4\left(x+y+z\right)}=\frac{\left(x+y+z\right)^2+18\left(x+y+z\right)+81}{4\left(x+y+z\right)}\)
\(4Q\ge x+y+z+\frac{81}{x+y+z}+18\ge2\sqrt{\frac{81\left(x+y+z\right)}{x+y+z}}+18=36\)
\(\Rightarrow Q\ge9\Rightarrow Q_{min}=9\) khi \(x=y=z=3\) hay \(a=b=c=9\)
Uầy, chưa ngủ luôn ạ