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Lời giải:
Áp dụng BĐT AM-GM:
\(P=\sum \sqrt{\frac{ab}{c+ab}}=\sum \sqrt{\frac{ab}{c(a+b+c)+ab}}=\sum \sqrt{\frac{ab}{(c+a)(c+b)}}\)
\(\leq \sum \frac{1}{2}\left(\frac{a}{c+a}+\frac{b}{c+b}\right)=\frac{1}{2}\left(\frac{a+b}{a+b}+\frac{b+c}{b+c}+\frac{c+a}{c+a}\right)=\frac{3}{2}\)
Vậy $P_{\max}=\frac{3}{2}$ khi $a=b=c=\frac{1}{3}$
Thỏa mãn $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1$ hay $a+b+c=1$ vậy bạn?
\(\dfrac{ab}{\sqrt{ab+2c}}=\dfrac{ab}{\sqrt{ab+\left(a+b+c\right)c}}=\dfrac{ab}{\sqrt{\left(a+c\right)\left(b+c\right)}}=ab\cdot\sqrt{\dfrac{1}{a+b}\cdot\dfrac{1}{b+c}}\le ab\cdot\dfrac{1}{2}\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}\right)=\dfrac{1}{2}\left(\dfrac{ab}{a+b}+\dfrac{ab}{b+c}\right)\)
CMTT: \(\dfrac{bc}{\sqrt{bc+2a}}\le\dfrac{1}{2}\left(\dfrac{bc}{a+b}+\dfrac{bc}{a+c}\right);\dfrac{ac}{\sqrt{ac+2b}}\le\dfrac{1}{2}\left(\dfrac{ac}{b+c}+\dfrac{ac}{b+a}\right)\)
\(\Leftrightarrow P\le\dfrac{1}{2}\left(\dfrac{ab}{c+a}+\dfrac{ab}{c+b}+\dfrac{bc}{b+a}+\dfrac{bc}{c+a}+\dfrac{ac}{b+c}+\dfrac{ac}{b+c}\right)\\ \Leftrightarrow P\le\dfrac{1}{2}\left[\dfrac{b\left(a+c\right)}{a+c}+\dfrac{a\left(b+c\right)}{b+c}+\dfrac{c\left(a+b\right)}{a+b}\right]=\dfrac{1}{2}\left(a+b+c\right)=1\)
Dấu \("="\Leftrightarrow a=b=c=\dfrac{2}{3}\)
Ta có: \(\frac{bc}{\sqrt{a+bc}}=\frac{bc}{\sqrt{\left(a+b\right)\left(a+c\right)}}\le\frac{\frac{bc}{a+b}+\frac{bc}{a+c}}{2}\)
\(\frac{ca}{\sqrt{b+ac}}=\frac{ca}{\sqrt{\left(a+b\right)\left(b+c\right)}}\le\frac{\frac{ca}{a+b}+\frac{ca}{b+c}}{2}\)
\(\frac{ab}{\sqrt{c+ab}}=\frac{ab}{\sqrt{\left(a+c\right)\left(b+c\right)}}\le\frac{\frac{ab}{a+c}+\frac{ab}{b+c}}{2}\)
Cộng 3 vế ta được: \(P\le\frac{\frac{bc}{a+b}+\frac{bc}{a+c}+\frac{ca}{a+b}+\frac{ca}{b+c}+\frac{ab}{a+c}+\frac{ab}{b+c}}{2}\)
\(=\frac{\frac{c\left(a+b\right)}{a+b}+\frac{b\left(a+c\right)}{a+c}+\frac{a\left(b+c\right)}{b+c}}{2}=\frac{a+b+c}{2}=\frac{1}{2}\)
Vậy MinP = 1/2
\(\frac{bc}{\sqrt{a+bc}}=\frac{bc}{\sqrt{a.1+bc}}=\frac{bc}{\sqrt{a\left(a+b+c\right)+bc}}=\frac{bc}{\sqrt{\left(a+b\right)\left(a+c\right)}}\le\frac{bc}{a+b}+\frac{bc}{a+c}\)
\(6a+3b+2c=abc\Leftrightarrow\dfrac{2}{ab}+\dfrac{3}{ac}+\dfrac{6}{bc}=1\)
Đặt \(\left(\dfrac{1}{a};\dfrac{2}{b};\dfrac{3}{c}\right)=\left(x;y;z\right)\Rightarrow xy+yz+zx=1\)
\(Q=\dfrac{1}{\sqrt{\dfrac{1}{x^2}+1}}+\dfrac{2}{\sqrt{\dfrac{4}{y^2}+4}}+\dfrac{3}{\sqrt{\dfrac{9}{z^2}+9}}=\dfrac{x}{\sqrt{x^2+1}}+\dfrac{y}{\sqrt{y^2+1}}+\dfrac{z}{\sqrt{z^2+1}}\)
\(Q=\dfrac{x}{\sqrt{x^2+xy+yz+zx}}+\dfrac{y}{\sqrt{y^2+xy+yz+zx}}+\dfrac{z}{\sqrt{z^2+xy+yz+zx}}\)
\(Q=\dfrac{x}{\sqrt{\left(x+y\right)\left(x+z\right)}}+\dfrac{y}{\sqrt{\left(x+y\right)\left(y+z\right)}}+\dfrac{z}{\sqrt{\left(x+z\right)\left(y+z\right)}}\)
\(Q\le\dfrac{1}{2}\left(\dfrac{x}{x+y}+\dfrac{x}{x+z}+\dfrac{y}{x+y}+\dfrac{y}{y+z}+\dfrac{z}{x+z}+\dfrac{z}{y+z}\right)=\dfrac{3}{2}\)
\(Q_{max}=\dfrac{3}{2}\) khi \(x=y=z=\dfrac{1}{\sqrt{3}}\) hay \(\left(a;b;c\right)=\left(\sqrt{3};2\sqrt{3};3\sqrt{3}\right)\)
\(a^2+b^2-ab\ge\dfrac{1}{2}\left(a+b\right)^2-\dfrac{1}{4}\left(a+b\right)^2=\dfrac{1}{4}\left(a+b\right)^2\)
\(\Rightarrow\dfrac{1}{\sqrt{a^2-ab+b^2}}\le\dfrac{1}{\sqrt{\dfrac{1}{4}\left(a+b\right)^2}}=\dfrac{2}{a+b}\le\dfrac{1}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\)
Tương tự:
\(\dfrac{1}{\sqrt{b^2-bc+c^2}}\le\dfrac{1}{2}\left(\dfrac{1}{b}+\dfrac{1}{c}\right)\) ; \(\dfrac{1}{\sqrt{c^2-ca+a^2}}\le\dfrac{1}{2}\left(\dfrac{1}{c}+\dfrac{1}{a}\right)\)
Cộng vế:
\(P\le\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=3\)
Dấu "=" xảy ra khi \(a=b=c=1\)
Có bất đẳng thức xy+zt≥x+zy+txy+zt≥x+zy+t với x,z≥0x,z≥0 ,y,t>0y,t>0
Giả sử cc lớn nhất trong các số a,b,ca,b,c thì c≥13c≥13
Do a,b,c≥0a,b,c≥0 nên
Ta có P2≥aa+1+bb+1+cc+1≥a+ba+b+2+cc+1P2≥aa+1+bb+1+cc+1≥a+ba+b+2+cc+1
Mà a+ba+b+2+cc+1−12=1−c3−c+c−12(c+1)=(1−c)(3c−1)(3−c)(2c+2)≥0
Đặt \(x=\sqrt{bc};y=\sqrt{ca};z=\sqrt{ab}\)\(\Rightarrow x^2+y^2+z^2+xyz=4\)\(\Rightarrow\left(x+y+z\right)^2-4=2\left(xy+yz+zx\right)-xyz\)
\(\Rightarrow\left(x+y+z\right)^2-4\left(x+y-z\right)+4=\left(2-x\right)\left(2-y\right)\left(2-z\right)\)\(\le\left(\frac{6-x-y-z}{3}\right)^3\)
Đặt \(t=x+y+z\Rightarrow\left(t-6\right)^3+27\left(t^2-4t+4\right)\le0\)\(\Leftrightarrow\left(t-3\right)\left(t+6\right)^2\le0\Leftrightarrow\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\le3\left(đpcm\right)\)
Dấu '=' xảy ra <=> a=b=c=1
Mình chưa hiểu ở dòng thứ 3 tại sao bạn lại đánh giá đc nó nhỏ hơn hoặc bằng \(\left(\frac{6-x-y-z}{3}\right)^3\)
\(P=\dfrac{\sqrt{a-1}}{a}+\dfrac{\sqrt{b-4}}{b}+\dfrac{\sqrt{c-9}}{c}=\dfrac{1.\sqrt{a-1}}{a}+\dfrac{2.\sqrt{b-4}}{2b}+\dfrac{3.\sqrt{c-9}}{3c}\)
Áp dụng hằng đẳng thức \(xy\le\dfrac{x^2+y^2}{2}\) ta được
\(P\le\dfrac{1+a-1}{2a}+\dfrac{4+b-4}{4b}+\dfrac{9+c-9}{6c}=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}=\dfrac{11}{12}\)
\(\Rightarrow P_{max}=\dfrac{11}{12}\) khi \(\left\{{}\begin{matrix}\sqrt{a-1}=1\\\sqrt{b-4}=2\\\sqrt{c-9}=3\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=2\\b=8\\c=18\end{matrix}\right.\)