Tham khảo:

Với các số thực không âm a,b,c thỏa mãn \(a^2+b^2+c^2=1\), tìm giá trị lớn nhất, giá trị nhỏ nhất của biểu thức:  \(Q=\s... - Hoc24

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 \(=\frac{3}{4}\)<=> \(a=b=c\)

Giúp mình với mình tick cho

Ta có \(\sqrt{1+a^2}+\sqrt{2a}\le\sqrt{2\left(1+a^2+2a\right)}=\sqrt{2}\left(a+1\right)\).

Tương tự \(\sqrt{1+b^2}+\sqrt{2b}\le\sqrt{2}\left(b+1\right)\); \(\sqrt{1+c^2}+\sqrt{2c}\le\sqrt{2}\left(c+1\right)\).

Lại có \(\left(2-\sqrt{2}\right)\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)\le\left(2-\sqrt{2}\right)\sqrt{3\left(a+b+c\right)}\le3\left(2-\sqrt{2}\right)\).

Do đó \(B\le\sqrt{2}\left(a+b+c+3\right)+3\left(2-\sqrt{2}\right)\le6\sqrt{2}+6-3\sqrt{2}=3\sqrt{2}+6\).

Dấu "=" xảy ra khi a = b = c = 1.

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 \(a=b=c=1\)

\(P=\dfrac{1}{2}\left(\dfrac{2\sqrt{bc}}{a+2\sqrt{bc}}+\dfrac{2\sqrt{ac}}{b+2\sqrt{ac}}+\dfrac{2\sqrt{ab}}{c+2\sqrt{ab}}\right)\)

\(P=\dfrac{1}{2}\left(1-\dfrac{a}{a+2\sqrt{bc}}+1-\dfrac{b}{b+2\sqrt{ca}}+1-\dfrac{c}{c+2\sqrt{ab}}\right)\)

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

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

\(P_{max}=1\) khi \(a=b=c\)

\(P^2=a+b+c+a^2+b^2+c^2+2\sqrt{\left(a+b^2\right)\left(b+c^2\right)}+2\sqrt{\left(b+c^2\right)\left(c+a^2\right)}+2\sqrt{\left(a+b^2\right)\left(c+a^2\right)}.\)

Theo bđt Bunhiacopski ta có

\(2\sqrt{\left(a+b^2\right)\left(b+c^2\right)}\ge2\sqrt{b^3}\)(vì \(a,c\ge0\))

Tương tự \(2\sqrt{\left(b+c^2\right)\left(c+a^2\right)}\ge2\sqrt{c^3}\)

                \(2\sqrt{\left(c+a^2\right)\left(a+b^2\right)}\ge2\sqrt{a^3}\)

\(\Rightarrow P^2\ge a+b+c+a^2+b^2+c^2+2\sqrt{a^3}+2\sqrt{b^3}+2\sqrt{c^3}\)

Theo gt : \(\hept{\begin{cases}a,b,c\ge0\\a^2+b^2+c^2=1\end{cases}\Rightarrow0\le a,b,c\le1}\)

\(\Rightarrow\hept{\begin{cases}a\ge a^2,b\ge b^2,c\ge c^2\\a^3\ge a^4,b^3\ge b^4,c^3\ge c^4\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}a+b+c\ge a^2+b^2+c^2=1\\2\sqrt{a^3}+2\sqrt{b^3}+2\sqrt{c^3}\ge2\left(a^2+b^2+c^2\right)=2\end{cases}}\)

\(\Rightarrow P^2\ge1+1+2=4\)\(\Rightarrow P\ge2\)

Dấu "=" xảy ra khi a=b=0,c=1 và các hoán vị của nó

Tìm Max

Theo bđt Bunhiacopski ta có

\(P^2\le\left(1+1+1\right)\left(a+b+c+a^2+b^2+c^2\right)\)

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

      \(=3\left(1+\sqrt{3}\right)\)

\(\Rightarrow P\le\sqrt{3\left(1+\sqrt{3}\right)}\)

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