1+1 =2 đó 

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ta là nhà tiên chi đây

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chắc chắn bọ̣̣̣̣̣̣̣̣̣n mày sẽ̃̃̃̃̃̃̃̃ nhấ́́́́́n đọc thêm

Đặt \(\left(\frac{1}{x};\frac{1}{y}\right)=\left(a;b\right)\Rightarrow ab+a+b=3\)

\(\Rightarrow ab+2\sqrt{ab}\le3\Rightarrow\left(\sqrt{ab}+3\right)\left(\sqrt{ab}-1\right)\le0\)

\(\Rightarrow\sqrt{ab}\le1\Rightarrow ab\le1\)

\(P=\frac{a}{\sqrt{3+a^2}}+\frac{b}{\sqrt{3+b^2}}=\frac{a}{\sqrt{ab+a+b+a^2}}+\frac{b}{\sqrt{ab+a+b+b^2}}\)

\(=\frac{a}{\sqrt{\left(a+b\right)\left(a+1\right)}}+\frac{b}{\sqrt{\left(a+b\right)\left(b+1\right)}}\le\frac{1}{2}\left(\frac{a}{a+b}+\frac{a}{a+1}+\frac{b}{a+b}+\frac{b}{b+1}\right)\)

\(P\le\frac{1}{2}\left(1+\frac{a}{a+1}+\frac{b}{b+1}\right)=\frac{1}{2}\left(1+\frac{ab+a+ab+b}{ab+a+b+1}\right)=\frac{1}{2}\left(1+\frac{ab+3}{4}\right)\)

\(P\le\frac{1}{2}\left(1+\frac{1+3}{4}\right)=1\)

Dấu " = " xảy ra khi \(a=b=1\) hay \(x=y=1\)

Chúc bạn học tốt !!!

Đặt \(\left(\frac{1}{x};\frac{1}{y}\right)=\left(a;b\right)\Rightarrow ab+a+b=3\)

\(\Rightarrow ab+2\sqrt{ab}\le3\Rightarrow\left(\sqrt{ab}+3\right)\left(\sqrt{ab}-1\right)\le0\)

\(\Rightarrow\sqrt{ab}\le1\Rightarrow ab\le1\)

\(P=\frac{a}{\sqrt{3+a^2}}+\frac{b}{\sqrt{3+b^2}}=\frac{a}{\sqrt{ab+a+b+a^2}}+\frac{b}{\sqrt{ab+a+b+b^2}}\)

\(=\frac{a}{\sqrt{\left(a+b\right)\left(a+1\right)}}+\frac{b}{\sqrt{\left(a+b\right)\left(b+1\right)}}\le\frac{1}{2}\left(\frac{a}{a+b}+\frac{a}{a+1}+\frac{b}{a+b}+\frac{b}{b+1}\right)\)

\(P\le\frac{1}{2}\left(1+\frac{a}{a+1}+\frac{b}{b+1}\right)=\frac{1}{2}\left(1+\frac{ab+a+ab+b}{ab+a+b+1}\right)=\frac{1}{2}\left(1+\frac{ab+3}{4}\right)\)

\(P\le\frac{1}{2}\left(1+\frac{1+3}{4}\right)=1\)

Dấu "=" xảy ra khi \(a=b=1\) hay \(x=y=1\)

Bài 1 :

Từ \(\left(x+1\right)\left(y+1\right)=4xy\)

\(\Rightarrow\frac{x+1}{x}.\frac{y+1}{y}=4\Leftrightarrow\left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right)=4\)

Đặt \(a=\frac{1}{x};b=\frac{1}{y}\), ta có :

\(\left(1+a\right)\left(1+b\right)=4\Leftrightarrow3=a+b+ab\Leftrightarrow\left(\sqrt{a}-\sqrt{b}\right)^2+2\sqrt{ab}+ab\ge2\sqrt{ab}+ab\)

Từ đó \(ab\le1\)

Áp dụng AM - GM cho 2 số thực dương ta có :
\(\frac{1}{\sqrt{3x^2+1}}=\frac{\frac{1}{x}}{\sqrt{3+\frac{1}{x^2}}}=\frac{a}{\sqrt{a+b+ab+a^2}}=\frac{a}{\sqrt{\left(a+b\right)\left(a+1\right)}}\le\frac{1}{2}\left(\frac{a}{a+b}+\frac{a}{a+1}\right)\)

Tương tự ta có :

\(\frac{1}{\sqrt{3y^2+1}}\le\frac{1}{2}.\left(\frac{a}{a+b}+\frac{b}{b+1}\right)\)

Bài 1 :

Vì \(\frac{2}{b}=\frac{1}{a}+\frac{1}{c}\) nên \(b=\frac{2ac}{a+c}\)

Do đó : \(\frac{a+b}{2a-b}=\frac{a+\frac{2ac}{a+c}}{2a-\frac{2ac}{a+c}}=\frac{c^2+3ac}{2a^2}=\frac{a+3c}{2a}\)

Và : \(\frac{c+b}{2c-b}=\frac{c+\frac{2ac}{a+c}}{2c-\frac{2ac}{a+c}}=\frac{c^2+3ac}{2c^2}=\frac{c+3a}{2c}\)

Suy ra \(P=\frac{a+b}{2a-b}+\frac{c+b}{2c-b}=\frac{a+3c}{2a}+\frac{c+3a}{2c}=\frac{ac+3c^2+ac+3a^2}{2ac}\)

\(=\frac{3\left(a^2+c^2\right)+2ac}{2ac}\ge\frac{3.2ac+2ac}{2ac}=\frac{8ac}{2ac}=4\)

Vậy \(P\ge4\) với mọi a,b,c thỏa mãn đề bài. Dấu bằng xảy ra khi a=b=c

Vậy GTNN của P là 4 khi a=b=c

Chúc bạn học tốt !!

Mình nhầm tí nhé bạn dổi bài 1 thành bài 2 nhé ạ sorry bạn nhìu

\(\left(1.x+9.\frac{1}{y}\right)^2\le\left(1^2+9^2\right)\left(x^2+\frac{1}{y^2}\right)\Rightarrow\sqrt{x^2+\frac{1}{y^2}}\ge\frac{1}{\sqrt{82}}\left(x+\frac{9}{y}\right)\)

\(TT:\sqrt{y^2+\frac{1}{z^2}}\ge\frac{1}{\sqrt{82}}\left(y+\frac{9}{z}\right);\sqrt{z^2+\frac{1}{x^2}}\ge\frac{1}{\sqrt{82}}\left(z+\frac{9}{x}\right)\)

\(S\ge\frac{1}{\sqrt{82}}\left(x+y+z+\frac{9}{x}+\frac{9}{y}+\frac{9}{z}\right)\ge\frac{1}{\sqrt{82}}\left(x+y+z+\frac{81}{x+y+z}\right)\)

\(=\frac{1}{\sqrt{82}}\left[\left(x+y+z+\frac{1}{x+y+z}\right)+\frac{80}{x+y+z}\right]\ge\sqrt{82}\)

cho bài cm hình đi

vd như Cho hình bình hành ABCD. trên các cạnh AB, BC, CD, DA theo thứ tự lấy các điểm M, N, P, Q sao cho AM = BN = CP = DQ. Chứng minh tứ giác MNPQ là hình bình hành

Chả biết đề có đúng không nữa nhưng mà nếu thử x = 0 ; y = -1 thì VT = 1,5 > 1 :)

Đặt \(\left(\frac{1}{x};\frac{1}{y}\right)=\left(a;b\right)\).Từ gt\(\Rightarrow\frac{\left(x+1\right)\left(y+1\right)}{xy}=4\Rightarrow\left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right)=4\)

\(\Rightarrow\left(a+1\right)\left(b+1\right)=4\Rightarrow ab+a+b=3\)

Mà \(ab+a+b=\left(\sqrt{a}-\sqrt{b}\right)^2+ab+2\sqrt{ab}\ge ab+2\sqrt{ab}\)

\(\Rightarrow3\ge ab+2\sqrt{ab}\Rightarrow4\ge\left(\sqrt{ab}+1\right)^2\)\(\Rightarrow\sqrt{ab}+1\le2\Rightarrow ab\le1\)

Ta có:\(\frac{1}{\sqrt{3x^2+1}}=\frac{\frac{1}{x}}{\sqrt{3+\frac{1}{x^2}}}=\frac{a}{\sqrt{ab+a+b+a^2}}=\frac{a}{\sqrt{\left(a+1\right)\left(a+b\right)}}\)\(\le\frac{1}{2}\left(\frac{a}{a+1}+\frac{a}{a+b}\right)\)

Tương tự:\(\frac{1}{\sqrt{3y^2+1}}\le\frac{1}{2}\left(\frac{b}{b+1}+\frac{b}{a+b}\right)\)

Cộng theo vế BĐT ta có:\(\frac{1}{\sqrt{3x^2+1}}+\frac{1}{\sqrt{3y^2+1}}\le\frac{1}{2}\left(\frac{a}{a+1}+\frac{a}{a+b}+\frac{b}{b+1}+\frac{b}{a+b}\right)\)

\(=\frac{1}{2}\left(\frac{a}{a+1}+\frac{b}{b+1}+1\right)=\frac{1}{2}\left[\frac{2ab+a+b}{\left(a+1\right)\left(b+1\right)}+1\right]\)\(\le\frac{1}{2}\left[\frac{ab+a+b+1}{\left(a+1\right)\left(b+1\right)}+1\right]=1\)
Dấu "=" xảy ra khi \(x=y=1\)

\(\left(x+1\right)\left(y+1\right)=4xy\Leftrightarrow\left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right)=4\)

Đặt \(\left(\frac{1}{x};\frac{1}{y}\right)=\left(a;b\right)\Rightarrow\left(a+1\right)\left(b+1\right)=4\Rightarrow ab+a+b=3\)

\(VT=\frac{a}{\sqrt{a^2+3}}+\frac{b}{\sqrt{b^2+3}}\)

Áp dụng BĐT Bunhiacopxki:

\(\left(a+\sqrt{3}.\sqrt{3}\right)^2\le\left(1+3\right)\left(a^2+3\right)\Rightarrow a^2+3\ge\frac{\left(a+3\right)^2}{4}\)

\(\Rightarrow VT\le\frac{2a}{a+3}+\frac{2b}{b+3}=\frac{4ab+6\left(a+b\right)}{ab+3\left(a+b\right)+9}=\frac{4\left(ab+a+b\right)+2\left(a+b\right)}{ab+a+b+9+2\left(a+b\right)}=\frac{12+2\left(a+b\right)}{12+2\left(a+b\right)}=1\)

Dấu "=" xảy ra khi \(a=b=1\) hay \(x=y=1\)