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xí câu 1:))
Áp dụng bất đẳng thức Cauchy-Schwarz dạng Engel ta có :
\(\frac{x^2}{y-1}+\frac{y^2}{x-1}\ge\frac{\left(x+y\right)^2}{x+y-2}\)(1)
Đặt a = x + y - 2 => a > 0 ( vì x,y > 1 )
Khi đó \(\left(1\right)=\frac{\left(a+2\right)^2}{a}=\frac{a^2+4a+4}{a}=\left(a+\frac{4}{a}\right)+4\ge2\sqrt{a\cdot\frac{4}{a}}+4=8\)( AM-GM )
Vậy ta có đpcm
Đẳng thức xảy ra <=> a=2 => x=y=2
Vi a^2+b^2+c^2=1
=>-1=<a,b,c=<1
=>(1+a)(1+b)(1+c)>=0
=>1+abc+ab+bc+ca+a+b+c>=0 (1*)
Lại có (a+b+c+1)^2/2>=0
=>[a^2+b^2+c^2+1+2a+2b+2c+2ab+2bc+2ca
]/2>=0
=>[2+2a+2b+2c+2ab+2bc+2ca]/2>=0 (Thay a^2+b^2+c^2=1)
=>1+a+b+c+ab+bc+ca>=0 (2*)
tu (1*)(2*) ta co abc+2(1+a+b+c+ab+bc+ca)>=0
dau = xay ra <=>a+b+c=-1 va a^2+b^2+c^2=1
<=>a=0,b=0,c=-1 va cac hoan vi cua no
Vì a^2+b^2+c^2=1
=>-1=<a,b,c=<1
=>(1+a)(1+b)(1+c)>=0
=>1+abc+ab+bc+ca+a+b+c>=0 (1*)
Lại có (a+b+c+1)^2/2>=0
=>[a^2+b^2+c^2+1+2a+2b+2c+2ab+2bc+2ca
]/2>=0
=>[2+2a+2b+2c+2ab+2bc+2ca]/2>=0 (Thay a^2+b^2+c^2=1)
=>1+a+b+c+ab+bc+ca>=0 (2*)
tu (1*)(2*) ta co abc+2(1+a+b+c+ab+bc+ca)>=0
dau = xay ra <=>a+b+c=-1 va a^2+b^2+c^2=1
<=>a=0,b=0,c=-1 và các hoan vi của nó
Đặt \(\left(x;y;z\right)=\left(\frac{1}{a};\frac{1}{b};\frac{1}{c}\right)\) \(\left(x,y,z>0\right)\)
Theo đề \(ab+bc+ca=3abc\Leftrightarrow\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}=\frac{3}{xyz}\)
\(\Rightarrow x+y+z=3\)
Và \(\sqrt{\frac{ab}{a+b+1}}+\sqrt{\frac{bc}{b+c+1}}+\sqrt{\frac{ca}{c+a+1}}\)
\(=\sqrt{\frac{\frac{1}{xy}}{\frac{1}{x}+\frac{1}{y}+1}}+\sqrt{\frac{\frac{1}{yz}}{\frac{1}{y}+\frac{1}{z}+1}}+\sqrt{\frac{\frac{1}{zx}}{\frac{1}{z}+\frac{1}{x}+1}}\)
\(=\frac{1}{\sqrt{x+y+xy}}+\frac{1}{\sqrt{y+z+yz}}+\frac{1}{\sqrt{z+x+zx}}\)
\(\ge\frac{9}{\sqrt{x+y+xy}+\sqrt{y+z+yz}+\sqrt{z+x+zx}}\) (Cauchy Schwarz)
Ta có: \(\sqrt{x+y+xy}+\sqrt{y+z+yz}+\sqrt{z+x+zx}\)
\(=\sqrt{\left(\sqrt{x+y+xy}+\sqrt{y+z+yz}+\sqrt{z+x+zx}\right)^2}\)
\(\le\sqrt{3\left(x+y+xy+y+z+yz+z+x+zx\right)}\)
\(=\sqrt{\left[2\left(x+y+z\right)+\left(xy+yz+zx\right)\right]}\)
\(\le\sqrt{6+\frac{\left(x+y+z\right)^2}{3}}=\sqrt{6+\frac{3^2}{3}}=3\)
\(\Rightarrow\sqrt{\frac{ab}{a+b+1}}+\sqrt{\frac{bc}{b+c+1}}+\sqrt{\frac{ca}{c+a+1}}\)
\(\ge\frac{9}{\sqrt{x+y+xy}+\sqrt{y+z+yz}+\sqrt{z+x+zx}}\ge\frac{9}{3}=3\)
Dấu "=" xảy ra khi: \(x=y=z=1\Rightarrow a=b=c=1\)
đặt \(\sqrt{\frac{ab}{c}}=x;\sqrt{\frac{bc}{a}}=y;\sqrt{\frac{ca}{b}}=z\Rightarrow xy+yz+zx=1\)
\(P=\frac{ab}{ab+c}+\frac{bc}{bc+a}+\frac{ca}{ca+b}\)
\(=\frac{\frac{ab}{c}}{\frac{ab}{c}+1}+\frac{\frac{bc}{a}}{\frac{bc}{a}+1}+\frac{\frac{ca}{b}}{\frac{ca}{b}+1}=\frac{x^2}{x^2+1}+\frac{y^2}{y^2+1}+\frac{z^2}{z^2+1}\)
\(\ge\frac{\left(x+y+z\right)^2}{\left(x+y+z\right)^2+\frac{\left(x+y+z\right)^2}{3}}=\frac{3}{4}\left(Q.E.D\right)\)
Ta có: \(\frac{a}{1+b^2}=\frac{a\left(1+b^2\right)-ab^2}{1+b^2}=a-\frac{ab}{1+b^2}\)
\(1+b^2\ge2b\) \(\Rightarrow\frac{ab^2}{1+b^2}\le\frac{ab^2}{2b}=\frac{ab}{2}\)\(\Rightarrow-\frac{ab^2}{1+b^2}\ge-\frac{ab}{2}\)
Do đó: \(\frac{a}{1+b^2}=a-\frac{ab^2}{1+b^2}\ge a-\frac{ab}{2}\)
Tương tự: \(\frac{b}{1+c^2}\ge b-\frac{bc}{2}\); \(\frac{c}{1+a^2}\ge c-\frac{ca}{2}\)
Suy ra \(\frac{a}{1+b^2}+\frac{b}{1+c^2}+\frac{c}{1+a^2}+\frac{ab+bc+ca}{2}\ge a+b+c\)
Mặt khác ta có: \(3\ge\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{9}{a+b+c}\Rightarrow\frac{3}{a+b+c}\le1\)
\(\Rightarrow a+b+c\ge3\)
Do đó; \(\frac{a}{1+b^2}+\frac{b}{1+c^2}+\frac{c}{1+a^2}+\frac{ab+bc+ca}{2}\ge a+b+c\ge3\)(đpcm)
Dấu "=" xảy ra khi và chỉ khi \(a=b=c=1\)
tao loa