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\(a,b,c\ge0,a+b+c=1\Rightarrow0\le a,b,c\le1\)
\(đi\) \(cminh:\sqrt{3a+1}\ge a+1\Leftrightarrow3a+1-\left(a+1\right)^2\ge0\Leftrightarrow-a\left(a-1\right)\ge0\Leftrightarrow a\left(a-1\right)\le0\left(đúng\right)\)
\(tương\) \(tự\Rightarrow A\ge a+b+c+1+1+1=4\)
\(min=4\Leftrightarrow\left(a;b;c\right)=\left\{1,0,0\right\}\) \(hoán\) \(vị\)
Đặt biểu thức trên là \(A\)
Theo BĐT Bunhiacopxki ta có :
\(A=\sqrt{3a+1}+\sqrt{3b+1}+\sqrt{3c+1}\le\sqrt{\left(1+1+1\right)\left(3a+1+3b+1+3c+1\right)}\)
\(=\sqrt{9\left(a+b+c+1\right)}=\sqrt{9.4}=6\)
Dấu "=" xảy ra \(\Leftrightarrow a=b=c=1\)
Dặt x=a, y=2b,z=3c
Khi đó
\(P=\frac{yz}{\sqrt{x+yz}}+\frac{xz}{\sqrt{y+xz}}+\frac{xy}{\sqrt{z+xy}}\)và x+y+z=1
Ta có \(\frac{yz}{\sqrt{x+yz}}=\frac{yz}{\sqrt{x\left(x+y+z\right)+yz}}=\frac{yz}{\sqrt{\left(x+y\right)\left(x+z\right)}}\le\frac{1}{2}yz\left(\frac{1}{x+y}+\frac{1}{x+z}\right)\)
=> \(P\le\frac{1}{2}\left(\frac{xz}{x+y}+\frac{yz}{x+y}\right)+\frac{1}{2}\left(\frac{xy}{y+z}+\frac{xz}{y+z}\right)+...=\frac{1}{2}\left(x+y+z\right)\)
\(=\frac{1}{2}\)
Vậy \(MaxP=\frac{1}{2}\)khi x=y=z=1/3 hay \(\hept{\begin{cases}a=\frac{1}{3}\\b=\frac{1}{6}\\c=\frac{1}{9}\end{cases}}\)
\(P\ge\frac{1}{\sqrt{ab+bc+ca+c^2}}+\frac{1}{\sqrt{ab+bc+ca+c^2}}=\frac{2}{\sqrt{ab+bc+ca+c^2}}\)
\(P\ge\frac{2}{\sqrt{\left(a+c\right)\left(b+c\right)}}=\frac{4\sqrt{2}}{2\sqrt{\left(a+c\right)\left(2b+2c\right)}}\ge\frac{4\sqrt{2}}{a+c+2b+2c}=\sqrt{2}\)
\(P_{min}=\sqrt{2}\) khi \(\left(a;b;c\right)=\left(2;1;0\right)\)
\(S\ge3\frac{1}{\sqrt[6]{\left(a+2b+5c\right)\left(b+2c+5a\right)\left(c+2a+5b\right)}}.\)
\(S\ge\frac{3.4}{\sqrt[6]{\left(a+2b+5c\right)\left(b+2c+5a\right)\left(c+2a+5b\right).16.16.16}}\)
\(S\ge\frac{12}{\frac{a+2b+5c+b+2c+5a+c+2a+5b+16+16+16}{6}}\)
\(S\ge\frac{3}{4}\)
Dấu "=" xảy ra \(\Leftrightarrow a=b=c=2\)