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- Chứng minh \(\sqrt{2}\left(a+b+c\right)\le\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{c^2+a^2}\)
Ta có \(\sqrt{2}.\left(\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{c^2+a^2}\right)\ge2\left(a+b+c\right)\)
\(\Leftrightarrow\sqrt{2}\left(a+b+c\right)\le\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{c^2+a^2}\)
- Chứng minh \(\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{c^2+a^2}< \sqrt{3}\left(a+b+c\right)\)
Bạn chứng minh bằng biến đổi tương đương
1/ \(ab+bc+ac=3abc\Leftrightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=3\)
Ta có \(\frac{1}{a^2+1}+\frac{1}{b^2+1}+\frac{1}{c^2+1}\le\frac{1}{2a}+\frac{1}{2b}+\frac{1}{2c}=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=\frac{3}{2}\)
Vậy min P = 3/2 tại a = b = c = 1
\(a^2\sqrt{a}+b^2\sqrt{b}+c^2\sqrt{c}+\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}}\)
\(=\left(a^2\sqrt{a}+\frac{1}{\sqrt{a}}\right)+\left(b^2\sqrt{b}+\frac{1}{\sqrt{b}}\right)+\left(c^2\sqrt{c}+\frac{1}{\sqrt{c}}\right)\)
\(\ge2a+2b+2c\ge6\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2=6\)
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
Đặt \(\hept{\begin{cases}\sqrt{a^2+b^2}=x\\\sqrt{b^2+c^2}=y\\\sqrt{c^2+a^2}=z\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x,y,z>0\\x+y+z=1\end{cases}}\)
Và \(\hept{\begin{cases}a^2=\frac{x^2+z^2-y^2}{2}\\b^2=\frac{x^2+y^2-z^2}{2}\\c^2=\frac{y^2+z^2-x^2}{2}\end{cases}}\) và \(\hept{\begin{cases}b+c\le\sqrt{2\left(b^2+c^2\right)}=\sqrt{2}y\\a+b\le\sqrt{2}x\\c+a\le\sqrt{2}z\end{cases}}\)
\(\Rightarrow VT\ge\frac{1}{2\sqrt{2}}\left(\frac{x^2+z^2-y^2}{y}+\frac{x^2+y^2-z^2}{2z}+\frac{y^2+z^2-x^2}{x}\right)\)
\(\ge\frac{1}{2\sqrt{2}}\left(\frac{2\left(x+y+z\right)^2}{x+y+z}-\left(x+y+z\right)\right)\)
\(=\frac{1}{2\sqrt{2}}\left(x+y+z\right)=\frac{1}{2\sqrt{2}}\)
Ta có: \(\left(a^2+1\right)\left(\frac{1}{3}+1\right)\ge\left(\frac{a}{\sqrt{3}}+1\right)^2\)
\(\Rightarrow a^2+1\ge\frac{3}{4}\left(\frac{a}{\sqrt{3}}+1\right)^2\Rightarrow\sqrt{a^2+1}\ge\frac{\sqrt{3}}{2}\left(\frac{a}{\sqrt{3}}+1\right)=\frac{1}{2}\left(a+\sqrt{3}\right)\)
\(\Rightarrow M\le\sum\frac{2a}{a+\sqrt{3}}=\sum\frac{2a}{a+\frac{\sqrt{3}}{3}+\frac{\sqrt{3}}{3}+\frac{\sqrt{3}}{3}}\)
\(\Rightarrow M\le\frac{1}{8}\sum a\left(\frac{1}{a}+3\sqrt{3}\right)=\frac{3}{8}+\frac{3\sqrt{3}}{8}\left(a+b+c\right)\le\frac{3}{8}+\frac{3\sqrt{3}}{8}.\sqrt{3}=\frac{3}{2}\)
\(\Rightarrow M_{max}=\frac{3}{2}\) khi \(a=b=c=\frac{1}{\sqrt{3}}\)
1.Ta có: \(c+ab=\left(a+b+c\right)c+ab\)
\(=ac+bc+c^2+ab\)
\(=a\left(b+c\right)+c\left(b+c\right)\)
\(=\left(b+c\right)\left(a+b\right)\)
CMTT \(a+bc=\left(c+a\right)\left(b+c\right)\)
\(b+ca=\left(b+c\right)\left(a+b\right)\)
Từ đó \(P=\sqrt{\frac{ab}{\left(a+b\right)\left(b+c\right)}}+\sqrt{\frac{bc}{\left(c+a\right)\left(a+b\right)}}+\sqrt{\frac{ca}{\left(b+c\right)\left(a+b\right)}}\)
Ta có: \(\sqrt{\frac{ab}{\left(a+b\right)\left(b+c\right)}}\le\frac{1}{2}\left(\frac{a}{a+b}+\frac{b}{b+c}\right)\)( theo BĐT AM-GM)
CMTT\(\Rightarrow P\le\frac{1}{2}\left(\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{a+c}+\frac{b}{a+b}+\frac{c}{b+c}+\frac{a}{a+b}\right)\)
\(\Rightarrow P\le\frac{1}{2}.3\)
\(\Rightarrow P\le\frac{3}{2}\)
Dấu"="xảy ra \(\Leftrightarrow a=b=c\)
Vậy /...
\(\frac{a+1}{b^2+1}=a+1-\frac{ab^2-b^2}{b^2+1}=a+1-\frac{b^2\left(a+1\right)}{b^2+1}\ge a+1-\frac{b^2\left(a+1\right)}{2b}\)
\(=a+1-\frac{b\left(a+1\right)}{2}=a+1-\frac{ab+b}{2}\)
Tương tự rồi cộng lại:
\(RHS\ge a+b+c+3-\frac{ab+bc+ca+a+b+c}{2}\)
\(\ge a+b+c+3-\frac{\frac{\left(a+b+c\right)^2}{3}+a+b+c}{2}=3\)
Dấu "=" xảy ra tại \(a=b=c=1\)
ta có:
\(A^2=\left(\frac{a}{\sqrt{a^2+1}}+\frac{b}{\sqrt{b^2+1}}+\frac{c}{\sqrt{c^2+1}}\right)^2\le\left(a+b+c\right)\left(\frac{a}{a^2+1}+\frac{b}{b^2+1}+\frac{c}{c^2+1}\right)\) (BĐT Bu-nhi-a)
=>\(A^2\le\sqrt{3}\left(\frac{a}{a^2+1}+\frac{b}{b^2+1}+\frac{c}{c^2+1}\right)\) (*)
mặt khác ta có: \(a^2+1\ge2a\) (BĐT cauchy ) =>\(\frac{a}{a^2+1}\le\frac{1}{2}\)
tương tự ta có: \(\frac{b}{b^2+1}\le\frac{1}{2}\) ; \(\frac{c}{c^2+1}\le\frac{1}{2}\)
=> \(\frac{a}{a^2+1}+\frac{b}{b^2+1}+\frac{c}{c^2+1}\le\frac{1}{2}+\frac{1}{2}+\frac{1}{2}=\frac{3}{2}\) (**)
từ (*),(**) => \(A^2\le\sqrt{3}.\frac{3}{2}=\frac{3\sqrt{3}}{2}\)
=>\(A\le\sqrt{\frac{3\sqrt{3}}{2}}\)
=> GTLN của A là \(\sqrt{\frac{3\sqrt{3}}{2}}\) <=> a=b=c<\(\frac{\sqrt{3}}{3}\)
Ta có:
\(\frac{a}{\sqrt{a^2+1}}=\frac{a}{\sqrt{a^2+\frac{1}{3}+\frac{1}{3}+\frac{1}{3}}}\)
\(\le\frac{\sqrt[8]{27}a}{\sqrt{4\sqrt[4]{a^2}}}=\frac{\sqrt[8]{27a^6}}{2}\)
\(=\frac{\sqrt{3}}{2}.\sqrt[8]{a^6.\frac{1}{3}}\)
\(\le\frac{\sqrt{3}}{2}.\frac{6a+\frac{2}{\sqrt{3}}}{8}\left(1\right)\)
Tương tự ta cũng có:
\(\hept{\begin{cases}\frac{b}{\sqrt{b^2+1}}\le\frac{\sqrt{3}}{2}.\frac{6b+\frac{2}{\sqrt{3}}}{8}\left(2\right)\\\frac{c}{\sqrt{c^2+1}}\le\frac{\sqrt{3}}{2}.\frac{6c+\frac{2}{\sqrt{3}}}{8}\left(3\right)\end{cases}}\)
Từ (1), (2), (3)
\(\Rightarrow A\le\frac{\sqrt{3}}{2}.\left(\frac{6}{8\sqrt{3}}+\frac{6}{8}\left(a+b+c\right)\right)\)
\(\le\frac{\sqrt{3}}{2}.\left(\frac{3}{4\sqrt{3}}+\frac{3\sqrt{3}}{4}\right)=\frac{3}{2}\)
Dấu = xảy ra khi \(a=b=c=\frac{1}{\sqrt{3}}\)