ĐỊT MẸ

helloo

Ta có \(1+x^2=x^2+xy+yz+xz=\left(x+z\right)\left(x+y\right)\)

Khi đó BĐT <=>

 \(\frac{1}{\left(x+y\right)\left(x+z\right)}+\frac{1}{\left(y+z\right)\left(x+z\right)}+\frac{1}{\left(x+y\right)\left(y+z\right)}\ge\frac{2}{3}\left(\frac{x}{\sqrt{\left(x+z\right)\left(x+y\right)}}+...\right)\)

<=> \(\frac{x+y+z}{\left(x+y\right)\left(y+z\right)\left(x+z\right)}\ge\frac{1}{3}\left(\frac{x\sqrt{y+z}+y\sqrt{x+z}+z\sqrt{x+y}}{\sqrt{\left(x+y\right)\left(y+z\right)\left(x+z\right)}}\right)^3\)

<=>\(\left(x+y+z\right)\sqrt{\left(x+y\right)\left(x+z\right)\left(y+z\right)}\ge\frac{1}{3}\left(x\sqrt{y+z}+y\sqrt{x+z}+z\sqrt{x+y}\right)^3\)

<=> \(\left(x+y+z\right)\sqrt{\left(x+y\right)\left(y+z\right)\left(x+z\right)}\ge\frac{1}{3}\left(\sqrt{x\left(1-yz\right)}+\sqrt{y\left(1-xz\right)}+\sqrt{z\left(1-xy\right)}\right)^3\)(1)

Xét \(\left(x+y\right)\left(y+z\right)\left(x+z\right)\ge\frac{8}{9}\left(x+y+z\right)\left(xy+yz+xz\right)\)

<=> \(9\left[xy\left(x+y\right)+yz\left(y+z\right)+xz\left(x+z\right)+2xyz\right]\ge8\left(xy\left(x+y\right)+xz\left(x+z\right)+yz\left(y+z\right)+3xyz\right)\)

<=> \(xy\left(y+x\right)+yz\left(y+z\right)+xz\left(x+z\right)\ge6xyz\)

<=> \(x\left(y-z\right)^2+z\left(x-y\right)^2+y\left(x-z\right)^2\ge0\)luôn đúng

Khi đó (1) <=> 

\(\left(x+y+z\right).\frac{2\sqrt{2}}{3}.\sqrt{x+y+z}\ge\frac{1}{3}\left(\sqrt{x\left(1-yz\right)}+....\right)^3\) 

<=> \(\sqrt{2\left(x+y+z\right)}\ge\sqrt{x\left(1-yz\right)}+\sqrt{y\left(1-xz\right)}+\sqrt{z\left(1-xy\right)}\)

Áp dụng buniacopxki cho vế phải ta có 

\(\sqrt{x\left(1-yz\right)}+\sqrt{y\left(1-xz\right)}+\sqrt{z\left(1-xy\right)}\le\sqrt{\left(x+y+z\right)\left(3-xy-yz-xz\right)}\)

                                                                                                       \(=\sqrt{2\left(x+y+z\right)}\)

=> BĐT được CM

Dấu bằng xảy ra khi \(x=y=z=\frac{1}{\sqrt{3}}\)

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

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

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

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

Pt tương đương:

\(2\sqrt{3\left(x^2+y^2+z^2\right)}\ge\sqrt{xy}+\sqrt{yz}+\sqrt{xz}+3\)

Có: \(\sqrt{3\left(x^2+y^2+z^2\right)}\ge\sqrt{3\cdot3\left(xyz\right)^2}=3\)

Đồng thời:

\(\sqrt{xy}+\sqrt{yz}+\sqrt{xz}\le\frac{x+y}{2}+\frac{y+z}{2}+\frac{x+z}{2}=x+y+z\le\sqrt{\left(x+y+z\right)^2}\le\sqrt{3\left(x^2+y^2+z^2\right)}\)

Rồi cộng lại 

nhìn số 82 = 9² + 1 mà nghĩ ra p²

Ta có :

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

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

\(\Rightarrow\sqrt{x^2+\frac{1}{y^2}}+\sqrt{y^2+\frac{1}{z^2}}+\sqrt{z^2+\frac{1}{x^2}}\ge\frac{1}{\sqrt{82}}\left(x+y+z+\frac{9}{x}+\frac{9}{y}+\frac{9}{z}\right)\ge\frac{1}{\sqrt{82}}\left(x+y+z+\frac{81}{x+y+z}\right)\)

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

Chơi thêm một phát nữa cho phức tạp :D. Cách này em làm chơi thôi á! Dài dòng lắm!

Ta chứng minh BĐT phụ (hay còn gọi là Mincopxki): \(\sqrt{a^2+b^2}+\sqrt{c^2+d^2}\ge\sqrt{\left(a+c\right)^2+\left(b+d\right)^2}\) với a,b,c,d thuộc R.

Bình phương hai vế và khai triển ra hết:,ta có:\(BĐT\Leftrightarrow a^2+b^2+c^2+d^2+2\sqrt{\left(a^2+b^2\right)\left(c^2+d^2\right)}\ge a^2+b^2+c^2+d^2+2ac+2bd\)

\(\Leftrightarrow2\sqrt{\left(a^2+b^2\right)\left(c^2+d^2\right)}\ge2\left(ac+bd\right)\)

\(\Leftrightarrow\sqrt{\left(a^2+b^2\right)\left(c^2+d^2\right)}\ge\left(ac+bd\right)\)

Bình phương hai vế, BĐT tương đương với:

\(\left(a^2+b^2\right)\left(c^2+d^2\right)\ge\left(ac+bd\right)^2\). Nhân tung hết mấy cái ngoặc ra,ta cần c/m:

\(a^2c^2+a^2d^2+b^2c^2+b^2d^2\ge a^2c^2+b^2d^2+2abcd\)

\(\Leftrightarrow a^2d^2+b^2c^2\ge2.ad.bc\Leftrightarrow\left(ad-bc\right)^2\ge0\) (BĐT đúng)

Dấu "=" xảy ra khi \(ad=bc\).

Áp dụng BĐT trên hai lần,ta có:

\(VT\ge\sqrt{\left(x+y+z\right)^2+\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2}\)

\(\ge\sqrt{81\left(x+y+z\right)^2+\frac{81}{\left(x+y+z\right)^2}-80\left(x+y+z\right)^2}\)

\(\ge\sqrt{2\sqrt{81\left(x+y+z\right)^2.\frac{81}{\left(x+y+z\right)^2}}-80.1^2}\) (Cô si hay AM-GM các kiểu -__-")

\(=\sqrt{2.81-80}=\sqrt{82}\left(Q.E.D\right)\)

Dấu "=" xảy ra khi \(x=y=z=\frac{1}{3}\) (anh/chị giải rõ ra nha :( máy em bị lag rồi)

Theo giả thiết xy + yz + zx = 1 nên ta có: \(VT=\frac{1}{1+x^2}+\frac{1}{1+y^2}+\frac{1}{1+z^2}=\frac{1}{xy+yz+zx+x^2}+\frac{1}{xy+yz+zx+y^2}+\frac{1}{xy+yz+zx+z^2}=\frac{1}{\left(x+y\right)\left(x+z\right)}+\frac{1}{\left(y+x\right)\left(y+z\right)}+\frac{1}{\left(z+x\right)\left(z+y\right)}=\frac{2\left(x+y+z\right)}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}\)Theo bất đẳng thức Cauchy-Schwarz: \(\left(\frac{x}{\sqrt{1+x^2}}+\frac{y}{\sqrt{1+y^2}}+\frac{z}{\sqrt{1+z^2}}\right)^2\le\left(x+y+z\right)\left(\frac{x}{1+x^2}+\frac{y}{1+y^2}+\frac{z}{1+z^2}\right)=\left(x+y+z\right)\left(\frac{x}{\left(x+y\right)\left(x+z\right)}+\frac{y}{\left(y+z\right)\left(y+x\right)}+\frac{z}{\left(z+x\right)\left(z+y\right)}\right)=\frac{2\left(x+y+z\right)\left(xy+yz+zx\right)}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}=\frac{2\left(x+y+z\right)}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}\)\(\Rightarrow\frac{2}{3}\left(\frac{x}{\sqrt{1+x^2}}+\frac{y}{\sqrt{1+y^2}}+\frac{z}{\sqrt{1+z^2}}\right)^3\le\frac{4\left(x+y+z\right)}{3\left(x+y\right)\left(y+z\right)\left(z+x\right)}\left(\frac{x}{\sqrt{1+x^2}}+\frac{y}{\sqrt{1+y^2}}+\frac{z}{\sqrt{1+z^2}}\right)\)Ta cần chứng minh: \(\frac{2\left(x+y+z\right)}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}\ge\frac{4\left(x+y+z\right)}{3\left(x+y\right)\left(y+z\right)\left(z+x\right)}\left(\frac{x}{\sqrt{1+x^2}}+\frac{y}{\sqrt{1+y^2}}+\frac{z}{\sqrt{1+z^2}}\right)\)

hay \(\frac{x}{\sqrt{1+x^2}}+\frac{y}{\sqrt{1+y^2}}+\frac{z}{\sqrt{1+z^2}}\le\frac{3}{2}\)

Bất đẳng thức cuối đúng theo AM - GM do: \(\frac{x}{\sqrt{1+x^2}}+\frac{y}{\sqrt{1+y^2}}+\frac{z}{\sqrt{1+z^2}}=\sqrt{\frac{x}{x+y}.\frac{x}{x+z}}+\sqrt{\frac{y}{y+z}.\frac{y}{x+y}}+\sqrt{\frac{z}{z+x}.\frac{z}{z+y}}\le\frac{\left(\frac{x}{x+y}+\frac{x}{x+z}\right)+\left(\frac{y}{y+z}+\frac{y}{x+y}\right)+\left(\frac{z}{z+x}+\frac{z}{z+y}\right)}{2}=\frac{3}{2}\)Đẳng thức xảy ra khi \(x=y=z=\frac{1}{\sqrt{3}}\)

Đặt \(P=\frac{x}{\sqrt{1+x^2}}+\frac{y}{\sqrt{1+y^2}}+\frac{z}{\sqrt{1+z^2}}+\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\)

Do x,y,z là các số thực dương nên ta biến đổi \(P=\frac{1}{\sqrt{1+\frac{1}{x^2}}}+\frac{1}{\sqrt{1+\frac{1}{y^2}}}+\frac{1}{\sqrt{1+\frac{1}{z^2}}}+\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\)

Đặt \(a=\frac{1}{x^2};b=\frac{1}{y^2};c=\frac{1}{z^2}\left(a,b,c>0\right)\)thì \(xy+yz+zx=\frac{1}{\sqrt{ab}}+\frac{1}{\sqrt{bc}}+\frac{1}{\sqrt{ca}}=1\)và \(P=\frac{1}{\sqrt{1+a}}+\frac{1}{\sqrt{1+b}}+\frac{1}{\sqrt{1+c}}+a+b+c\)

Biến đổi biểu thức P=\(\left(\frac{1}{2\sqrt{a+1}}+\frac{1}{2\sqrt{a+1}}+\frac{a+1}{16}\right)+\left(\frac{1}{2\sqrt{b+1}}+\frac{1}{2\sqrt{b+1}}+\frac{b+1}{16}\right)\)\(+\left(\frac{1}{2\sqrt{c+1}}+\frac{1}{2\sqrt{c+1}}+\frac{c+1}{16}\right)+\frac{15a}{16}+\frac{15b}{16}+\frac{15c}{b}-\frac{3}{16}\)

Áp dụng Bất Đẳng Thức Cauchy ta có

\(P\ge3\sqrt[3]{\frac{a+1}{64\left(a+1\right)}}+3\sqrt[3]{\frac{b+1}{64\left(b+1\right)}}+3\sqrt[3]{\frac{c+1}{64\left(c+1\right)}}+\frac{15a}{16}+\frac{15b}{16}+\frac{15c}{16}-\frac{3}{16}\)

\(=\frac{33}{16}+\frac{15}{16}\left(a+b+c\right)\ge\frac{33}{16}+\frac{15}{16}\cdot3\sqrt[3]{abc}\)

Mặt khác ta có \(1=\frac{1}{\sqrt{ab}}+\frac{1}{\sqrt{bc}}+\frac{1}{\sqrt{ca}}\ge3\sqrt[3]{\frac{1}{abc}}\Leftrightarrow abc\ge27\)

\(\Rightarrow P\ge\frac{33}{16}+\frac{15}{16}\cdot3\sqrt[3]{27}=\frac{33}{16}+\frac{15}{16}\cdot9=\frac{21}{2}\)

Dấu "=" xảy ra khi a=b=c hay \(x=y=z=\frac{\sqrt{3}}{3}\)

Lời giải bài này khá dài, làm biếng gõ

Bạn lên google search "đề thi vào 10 chuyên khtn" nhé, đây là bài BĐT trong đề vòng 1 chuyên KHTN năm 2019

Ta có:

\( 1 + {x^2} = \left( {x + y} \right)\left( {x + z} \right)\\ 1 + {y^2} = \left( {x + y} \right)\left( {y + z} \right)\\ 1 + {z^2} = \left( {x + z} \right)\left( {y + z} \right) \)

Nên: \(\dfrac{1}{{1 + {x^2}}} + \dfrac{1}{{1 + {y^2}}} + \dfrac{1}{{1 + {z^2}}} = \dfrac{1}{{\left( {x + y} \right)\left( {x + z} \right)}} + \dfrac{1}{{\left( {x + y} \right)\left( {y + z} \right)}} + \dfrac{1}{{\left( {x + z} \right)\left( {y + z} \right)}} = \dfrac{{2\left( {x + y + z} \right)}}{{\left( {x + y} \right)\left( {y + z} \right)\left( {x + z} \right)}}\)

\( \dfrac{x}{{\sqrt {1 + {x^2}} }} + \dfrac{y}{{\sqrt {1 + {y^2}} }} + \dfrac{z}{{\sqrt {1 + {z^2}} }} = \dfrac{x}{{\sqrt {\left( {x + y} \right)\left( {x + z} \right)} }} + \dfrac{y}{{\sqrt {\left( {x + y} \right)\left( {y + z} \right)} }} + \dfrac{z}{{\left( {x + z} \right)\left( {y + z} \right)}}\\ \dfrac{x}{{\sqrt {1 + {x^2}} }} + \dfrac{y}{{\sqrt {1 + {y^2}} }} + \dfrac{z}{{\sqrt {1 + {z^2}} }} \le \dfrac{1}{2}\left( {\dfrac{x}{{x + y}} + \dfrac{x}{{x + z}} + \dfrac{y}{{x + y}} + \dfrac{y}{{y + z}} + \dfrac{z}{{x + z}} + \dfrac{z}{{y + z}}} \right) \)

Mặt khác, áp dụng $Bunhia$ ta có:

\({\left( {\dfrac{x}{{\sqrt {1 + {x^2}} }} + \dfrac{y}{{\sqrt {1 + {y^2}} }} + \dfrac{z}{{\sqrt {1 + {z^2}} }}} \right)^2} \le \left( {x + y + z} \right)\left( {\dfrac{x}{{1 + {x^2}}} + \dfrac{y}{{1 + {y^2}}} + \dfrac{z}{{1 + {z^2}}}} \right) = M\)

Với \(M = \dfrac{{2\left( {x + y + z} \right)\left( {xy + yz + xz} \right)}}{{\left( {x + y} \right)\left( {x + z} \right)\left( {y + z} \right)}} = \dfrac{{2\left( {x + y + z} \right)}}{{\left( {x + y} \right)\left( {x + z} \right)\left( {y + z} \right)}}\)

Lại có:

\( VP = \dfrac{2}{3}{\left( {\dfrac{x}{{1 + {x^2}}} + \dfrac{y}{{1 + {y^2}}} + \dfrac{z}{{1 + {z^2}}}} \right)^3} = \dfrac{2}{3}{\left( {\dfrac{1}{{1 + {x^2}}} + \dfrac{1}{{1 + {y^2}}} + \dfrac{1}{{1 + {z^2}}}} \right)^2}\\ VP \le \dfrac{{4\left( {x + y + z} \right)}}{{3\left( {x + y} \right)\left( {x + z} \right)\left( {y + z} \right)}}.\dfrac{3}{2} = \dfrac{{2\left( {x + y + z} \right)}}{{\left( {x + y} \right)\left( {x + z} \right)\left( {y + z} \right)}} = \dfrac{1}{{1 + {x^2}}} + \dfrac{1}{{1 + {y^2}}} + \dfrac{1}{{1 + {z^2}}} \)

Vậy \(\dfrac{1}{{1 + {x^2}}} + \dfrac{1}{{1 + {y^2}}} + \dfrac{1}{{1 + {z^2}}} \ge \dfrac{3}{2}{\left( {\dfrac{x}{{\sqrt {1 + {x^2}} }} + \dfrac{y}{{\sqrt {1 + {y^2}} }} + \dfrac{z}{{\sqrt {1 + {z^2}} }}} \right)^2}\)

Dấu \("= "\) xảy ra khi \(x=y=z=\dfrac{1}{\sqrt{3}}\)