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\(a^2-ab+b^2=\frac{1}{4}\left(a+b\right)^2+\frac{3}{4}\left(a-b\right)^2\ge\frac{1}{4}\left(a+b\right)^2\)
\(\frac{1}{\sqrt{a^2-ab+b^2}}\le\frac{1}{\sqrt{\frac{1}{4}\left(a+b\right)^2}}=\frac{2}{a+b}\le\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)\)
Tương tự với 2 số còn lại, cộng theo vế ta được kết quả cần tìm.
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
\(B=\frac{a}{\sqrt{1+a^2}}+\frac{b}{\sqrt{1+b^2}}+\frac{c}{\sqrt{1+c^2}}\)
\(B=\frac{a}{\sqrt{\left(a+b\right)\left(a+c\right)}}+\frac{b}{\sqrt{\left(a+b\right)\left(b+c\right)}}+\frac{c}{\sqrt{\left(b+c\right)\left(c+a\right)}}\)
\(B\le\frac{\frac{a}{a+b}+\frac{a}{a+c}+\frac{b}{a+b}+\frac{b}{b+c}+\frac{c}{b+c}+\frac{c}{c+a}}{2}=\frac{3}{2}\Rightarrow B_{max}=\frac{3}{2}\)
\(\text{Dấu "=" xảy ra khi và chỉ khi:}a=b=c=\frac{1}{\sqrt{3}}\)
\(A=\sqrt{2b\left(a+1\right)}+\sqrt{2c\left(b+1\right)}+\sqrt{2a\left(c+1\right)}\)
\(A=\dfrac{1}{2\sqrt{2}}.2\sqrt{4b\left(a+1\right)}+\dfrac{1}{2\sqrt{2}}.2\sqrt{4c\left(b+1\right)}+\dfrac{1}{2\sqrt{2}}.2\sqrt{4a\left(c+1\right)}\)
\(A\le\dfrac{1}{2\sqrt{2}}\left(4b+a+1\right)+\dfrac{1}{2\sqrt{2}}\left(4c+b+1\right)+\dfrac{1}{2\sqrt{2}}\left(4a+c+1\right)\)
\(A\le\dfrac{1}{2\sqrt{2}}\left[5\left(a+b+c\right)+3\right]=2\sqrt{2}\)
Dấu "=" xảy ra khi \(a=b=c=\dfrac{1}{3}\)
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 $a=b=c=\frac{1}{3}$
Cô-si : \(\sqrt{\left(a+b\right)\left(b+c\right)}\le\frac{a+b+b+c}{2}=\frac{a+2b+c}{2}\)
\(\frac{a}{\sqrt{bc\left(1+a^2\right)}}=\frac{a}{\sqrt{bc+a\left(a+b+c\right)}}=a\sqrt{\frac{1}{a+b}.\frac{1}{c+a}}\le\frac{\frac{a}{a+b}+\frac{a}{c+a}}{2}\)
Tương tự 2 cái còn lại cộng lại ta đc \(VT\le\frac{3}{2}\)
Dấu "=" xảy ra khi \(a=b=c=\sqrt{3}\)
Cach khac
Dat \(P=\frac{a}{\sqrt{bc\left(1+a^2\right)}}+\frac{b}{\sqrt{ca\left(1+b^2\right)}}+\frac{c}{\sqrt{ab\left(1+c^2\right)}}\)
Ta co:
\(a+b+c=abc\)
\(\Rightarrow\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=1\)
Dat \(\left(\frac{1}{a};\frac{1}{b};\frac{1}{c}\right)=\left(x;y;z\right)\)
\(\Rightarrow xy+yz+zx=1\)
\(\Rightarrow P=\sqrt{\frac{yz}{1+x^2}}+\sqrt{\frac{zx}{1+y^2}}+\sqrt{\frac{xy}{1+z^2}}\)
Ta lai co:
\(\sqrt{\frac{yz}{1+x^2}}=\sqrt{\frac{yz}{xy+yz+zx+x^2}}=\sqrt{\frac{yz}{\left(x+y\right)\left(z+x\right)}}\le\frac{1}{2}\left(\frac{y}{x+y}+\frac{z}{z+x}\right)\)
Tuong tu:
\(\sqrt{\frac{zx}{1+y^2}}\le\frac{1}{2}\left(\frac{z}{y+z}+\frac{x}{x+y}\right)\)
\(\sqrt{\frac{xy}{1+z^2}}\le\frac{1}{2}\left(\frac{x}{z+x}+\frac{y}{y+z}\right)\)
\(\Rightarrow P\le\frac{1}{2}\left(\frac{x+y}{x+y}+\frac{y+z}{y+z}+\frac{z+x}{z+x}\right)=\frac{3}{2}\)
Dau '=' xay ra khi \(x=y=z=\frac{1}{\sqrt{3}}\)
\(\Rightarrow a=b=c=\sqrt{3}\)
Vay \(P_{min}=\frac{3}{2}\)khi \(a=b=c=\sqrt{3}\)
+ Theo bđt Cauchy :
\(\sqrt{\left(a+b\right)\cdot\frac{2}{3}}\le\frac{a+b+\frac{2}{3}}{2}\) Dấu "=" \(\Leftrightarrow a+b=\frac{2}{3}\)
\(\sqrt{\left(b+c\right)\cdot\frac{2}{3}}\le\frac{b+c+\frac{2}{3}}{2}\) Dấu "=" \(\Leftrightarrow b+c=\frac{2}{3}\)
\(\sqrt{\left(c+a\right)\cdot\frac{2}{3}}\le\frac{c+a+\frac{2}{3}}{2}\) Dấu "=" \(\Leftrightarrow c+a=\frac{2}{3}\)
Do đó : \(\sqrt{\frac{2}{3}}Q\le\frac{2\left(a+b+c\right)+2}{2}=2\)
\(\Rightarrow Q\le\sqrt{6}\)
"=" \(\Leftrightarrow a=b=c=\frac{1}{3}\)