Bài này hay:)

c = min {a,b,c}. Đặt

\(a-c=x;b-c=y\Rightarrow x,y\ge0\) và x + y = a + b - 2c \(=3-3c\le3\)

\(\Rightarrow a-b=x-y;c=\frac{3-x-y}{3}\)

\(a=x+c=x+\frac{3-x-y}{3}=\frac{2x-y+3}{3}\)

\(b=y+c=\frac{2y-x+3}{3}\)

Như vậy: \(K=\sqrt{4\left(2x-y+3\right)+y^2}+\sqrt{4\left(2y-x+3\right)+x^2}+\sqrt{4\left(3-x-y\right)+\left(x-y\right)^2}\)

\(=\sqrt{y^2-4y+8x+12}+\sqrt{x^2-4x+8y+12}+\sqrt{4\left(3-x-y\right)+\left(x-y\right)^2}\)

Giờ em đang bận, tối em làm tiếp!

\(12a+\left(b-c\right)^2=4a\left(a+b+c\right)+b^2-2bc+c^2\)

\(=4a^2+b^2+c^2+4ab+4ac+2bc-4bc\)

\(=\left(2a+b+c\right)^2-4bc\le\left(2a+b+c\right)^2\)

\(\Rightarrow\sqrt{12a+\left(b-c\right)^2}\le2a+b+c\)

Tương tự: \(\sqrt{12b+\left(a-c\right)^2}\le a+2b+c\); \(\sqrt{12c+\left(a-b\right)^2}\le a+b+2c\)

Cộng vế với vế:

\(K\le4\left(a+b+c\right)=12\)

Dấu "=" xảy ra khi \(\left(a;b;c\right)=\left(0;0;3\right)\) và các hoán vị

Sử dụng AM-GM, ta có

\(P=\sum\sqrt{\dfrac{ab}{ab+c}}=\sum\sqrt{\dfrac{ab}{ab+c\left(a+b+c\right)}}=\sum\sqrt{\dfrac{ab}{\left(c+b\right)\left(c+a\right)}}\le\dfrac{1}{2}\sum\dfrac{a}{c+b}+\dfrac{b}{c+a}=\dfrac{3}{2}\)

Ta có: 

Theo bất đẳng thức Cô - si, ta có: \(\sqrt{\left(a+b\right)\left(a+c\right)}+\sqrt{bc}\le\frac{a+b+a+c}{2}+\frac{b+c}{2}=1\)

\(\Rightarrow\sqrt{a}\left(\sqrt{\left(a+b\right)\left(a+c\right)}+\sqrt{bc}\right)\le\sqrt{a}\)hay \(\sqrt{a^2+abc}+\sqrt{abc}\le\sqrt{a}\)

Tương tự ta có: \(\sqrt{b^2+abc}+\sqrt{abc}\le\sqrt{b}\);\(\sqrt{c^2+abc}+\sqrt{abc}\le\sqrt{c}\)

Mà \(abc\le\left(\frac{a+b+c}{3}\right)^3=\frac{1}{27}\Rightarrow\sqrt{abc}\le\frac{1}{3\sqrt{3}}\)

\(\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2\le3\left(a+b+c\right)=3\)\(\Leftrightarrow\sqrt{a}+\sqrt{b}+\sqrt{c}\le\sqrt{3}\)

Dấu "=" xảy ra khi và chỉ khi \(a=b=c=\frac{1}{3}\)

a=b=c=1/3

tôi đúng mừ

ko khó nhưng mà bn đăng từng câu 1 hộ mk mk giải giúp cho

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 <=> \(a=b=c=\sqrt{3}\) 

Đặt: \(\sqrt{a}=x;\sqrt{b}=y;\sqrt{c}=z\)

=>     \(P=\frac{xy}{z^2+3xy}+\frac{yz}{x^2+3yz}+\frac{zx}{y^2+3zx}\)

=>     \(3P=\frac{3xy}{z^2+3xy}+\frac{3yz}{x^2+3yz}+\frac{3zx}{y^2+3zx}=1-\frac{z^2}{z^2+3xy}+1-\frac{x^2}{x^2+3yz}+1-\frac{y^2}{y^2+3zx}\)

Ta sẽ CM: \(3P\le\frac{9}{4}\)<=> Cần CM: \(\frac{x^2}{x^2+3yz}+\frac{y^2}{y^2+3zx}+\frac{z^2}{z^2+3xy}\ge\frac{3}{4}\)

Có:    \(VT\ge\frac{\left(x+y+z\right)^2}{x^2+y^2+z^2+3\left(xy+yz+zx\right)}\)

Ta sẽ CM: \(\frac{\left(x+y+z\right)^2}{x^2+y^2+z^2+3\left(xy+yz+zx\right)}\ge\frac{3}{4}\)

<=> \(4\left(x+y+z\right)^2\ge3\left(x^2+y^2+z^2\right)+9\left(xy+yz+zx\right)\)

<=> \(4\left(x^2+y^2+z^2\right)+8\left(xy+yz+zx\right)\ge3\left(x^2+y^2+z^2\right)+9\left(xy+yz+zx\right)\)

<=> \(x^2+y^2+z^2\ge xy+yz+zx\)

Mà đây lại là 1 BĐT luôn đúng => \(3P\le\frac{9}{4}\)=> \(P\le\frac{3}{4}\)

Vậy P max \(=\frac{3}{4}\)<=> \(a=b=c\)

Ta có:

\(ab+bc+ca\le\dfrac{1}{3}\left(a+b+c\right)^2=3\)

\(\Rightarrow\dfrac{a}{\sqrt{a^2+3}}\le\dfrac{a}{\sqrt{a^2+ab+bc+ca}}=\dfrac{a}{\sqrt{\left(a+b\right)\left(a+c\right)}}\le\dfrac{1}{2}\left(\dfrac{a}{a+b}+\dfrac{a}{a+c}\right)\)

Tương tự:

\(\dfrac{b}{\sqrt{b^2+3}}\le\dfrac{1}{2}\left(\dfrac{b}{a+b}+\dfrac{b}{b+c}\right)\) ; \(\dfrac{c}{\sqrt{c^2+3}}\le\dfrac{1}{2}\left(\dfrac{c}{c+a}+\dfrac{c}{b+c}\right)\)

Cộng vế:

\(P\le\dfrac{1}{2}\left(\dfrac{a}{a+b}+\dfrac{b}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{b+c}+\dfrac{c}{a+c}+\dfrac{a}{a+c}\right)=\dfrac{3}{2}\)

\(P_{max}=\dfrac{3}{2}\) khi \(a=b=c=1\)