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Bài 1:
\(P=(x+1)\left(1+\frac{1}{y}\right)+(y+1)\left(1+\frac{1}{x}\right)\)
\(=2+x+y+\frac{x}{y}+\frac{y}{x}+\frac{1}{x}+\frac{1}{y}\)
Áp dụng BĐT Cô-si:
\(\frac{x}{y}+\frac{y}{x}\geq 2\)
\(x+\frac{1}{2x}\geq 2\sqrt{\frac{1}{2}}=\sqrt{2}\)
\(y+\frac{1}{2y}\geq 2\sqrt{\frac{1}{2}}=\sqrt{2}\)
Áp dụng BĐT SVac-xơ kết hợp với Cô-si:
\(\frac{1}{2x}+\frac{1}{2y}\geq \frac{4}{2x+2y}=\frac{2}{x+y}\geq \frac{2}{\sqrt{2(x^2+y^2)}}=\frac{2}{\sqrt{2}}=\sqrt{2}\)
Cộng các BĐT trên :
\(\Rightarrow P\geq 2+2+\sqrt{2}+\sqrt{2}+\sqrt{2}=4+3\sqrt{2}\)
Vậy \(P_{\min}=4+3\sqrt{2}\Leftrightarrow a=b=\frac{1}{\sqrt{2}}\)
Bài 2:
Áp dụng BĐT Svac-xơ:
\(\frac{1}{a+3b}+\frac{1}{b+a+2c}\geq \frac{4}{2a+4b+2c}=\frac{2}{a+2b+c}\)
\(\frac{1}{b+3c}+\frac{1}{b+c+2a}\geq \frac{4}{2b+4c+2a}=\frac{2}{b+2c+a}\)
\(\frac{1}{c+3a}+\frac{1}{c+a+2b}\geq \frac{4}{2c+4a+2b}=\frac{2}{c+2a+b}\)
Cộng theo vế và rút gọn :
\(\Rightarrow \frac{1}{a+3b}+\frac{1}{b+3c}+\frac{1}{c+3a}\geq \frac{1}{2a+b+c}+\frac{1}{2b+c+a}+\frac{1}{2c+a+b}\) (đpcm)
Dấu bằng xảy ra khi $a=b=c$
Dean thật, gõ gần xong rồi tự nhiên nó tạch, phải gõ lại -.-
Từ gt, ta suy ra:
\(a^3+b^3+c^3-3abc=0\)
\(\Leftrightarrow\left(a+b+c\right)\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right].\dfrac{1}{2}=0\)(Tự phân tích, không còn kiên nhẫn để gõ lại)
Mà a+b+c khác 0 => a=b=c
Thay vào thì C=8
bai 2 :
dat cac tich ab , bc , ca lan luot la x,y,z ( khac 0 )
thay vao ta dc : x^3+y^3+z^3=3xyz
=> (x+y)(x^2-2xy+y^2)+z^3-3xyz=0
=>(x+y)(x^2+2xy+y^2)+z^3-3xy(x+y)-3xyz=0
=》（x+y+z)【（x+y）^2 -(x+y)z+z^2】-3xy(x+y+z)=0
=>(x+y+z)(x^2+y^2+z^2-xy-yz-xz)=0
=>\(\dfrac{1}{2}\left(x+y+z\right)\left[\left(x-y\right)^2+\left(y-z\right)^2+\left(x-z\right)^2\right]\)=0
=> x+y+z=0 hoac x=y=z
TH1 : a+b+c=0
=>P=-1
TH2 : a=b=c
=>P=8
BĐT cơ bản
\(\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{4}{x+y}\)
\(\dfrac{ab}{c+1}=ab\dfrac{1}{c+a+b+c}=ab\dfrac{1}{\left(c+a\right)+\left(b+c\right)}\le\dfrac{ab}{4}\left[\dfrac{1}{c+a}+\dfrac{1}{b+c}\right]\)
\(\dfrac{bc}{a+1}=bc\dfrac{1}{a+a+b+c}=bc\dfrac{1}{\left(a+b\right)+\left(a+c\right)}\le\dfrac{bc}{4}\left[\dfrac{1}{a+b}+\dfrac{1}{a+c}\right]\)
\(\dfrac{ac}{b+1}=ac\dfrac{1}{b+a+b+c}=ac\dfrac{1}{\left(b+a\right)+\left(b+c\right)}\le\dfrac{ac}{4}\left[\dfrac{1}{b+a}+\dfrac{1}{b+c}\right]\)
Công lại:
\(A\le\left[\dfrac{ab+bc}{4\left(c+a\right)}+\dfrac{ab+ac}{4\left(b+c\right)}+\dfrac{bc+ac}{4\left(b+a\right)}\right]\)
\(A\le\left[\dfrac{b\left(a+c\right)}{4\left(c+a\right)}+\dfrac{a\left(b+c\right)}{4\left(b+c\right)}+\dfrac{c\left(b+a\right)}{4\left(b+a\right)}\right]\)
\(A\le\left[\dfrac{b}{4}+\dfrac{a}{4}+\dfrac{c}{4}\right]\)
\(A\le\dfrac{b+a+c}{4}=\dfrac{1}{4}\)
Đẳng thức khi \(a=b=c=\dfrac{1}{3}\)
Xong rồi đó mỏi cái lưng
Áp dụng BĐT Cauchy cho từng cặp số:
\(\dfrac{ab}{c+1}=\dfrac{bc}{a+1}\); \(\dfrac{bc}{a+1}=\dfrac{ca}{b+1}\) ; \(\dfrac{ac}{b+1}=\dfrac{ab}{c+1}\)
Kết quả cuối cùng là \(VT\ge a+b+c=1\)
Dấu " = " xảy ra khi và chỉ khi \(a=b=c=\dfrac{1}{3}\)
Không chắc :v
+ \(2a+b+c=\left(a+b\right)+\left(a+c\right)\)
\(\ge2\sqrt{\left(a+b\right)\left(a+c\right)}\) ( theo AM-GM )
\(\Rightarrow\left(2a+b+c\right)^2\ge4\left(a+b\right)\left(a+c\right)\)
\(\Rightarrow\frac{1}{\left(2a+b+c\right)^2}\le\frac{1}{4\left(a+b\right)\left(a+c\right)}\)
Dấu "=" xảy ra \(\Leftrightarrow b=c\)
+ Tương tự : \(\frac{1}{\left(2b+c+a\right)^2}\le\frac{1}{4\left(a+b\right)\left(b+c\right)}\). Dấu "=" xảy ra <=> a = c
\(\frac{1}{\left(2c+a+b\right)^2}\le\frac{1}{4\left(a+c\right)\left(b+c\right)}\). Dấu "=" xảy ra \(\Leftrightarrow a=b\)
Do đó : \(P\le\frac{1}{4}\left(\frac{1}{\left(a+b\right)\left(a+c\right)}+\frac{1}{\left(a+b\right)\left(b+c\right)}+\frac{1}{\left(a+c\right)\left(b+c\right)}\right)\)
\(\Rightarrow P\le\frac{1}{2}\cdot\frac{a+b+c}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)
\(\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge2\sqrt{ab}\cdot2\sqrt{bc}\cdot2\sqrt{ca}\)\(=8abc\)
\(\Rightarrow P\le\frac{a+b+c}{16abc}\)
+ \(\frac{1}{a^2}+\frac{1}{b^2}\ge\frac{2}{ab}\). Dấu :=" xảy ra \(\Leftrightarrow a=b\)
\(\frac{1}{b^2}+\frac{1}{c^2}\ge\frac{2}{bc}\). Dấu "=" xảy ra <=> b = c
\(\frac{1}{c^2}+\frac{1}{a^2}\ge\frac{2}{ca}\). Dấu "=" xảy ra <=> c = a
\(\Rightarrow2\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\ge2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)\)
\(\Rightarrow3\ge\frac{a+b+c}{abc}\) \(\Rightarrow a+b+c\le3abc\)
\(\Rightarrow P\le\frac{3abc}{16abc}=\frac{3}{16}\)
Dấu "=" xảy ra \(\Leftrightarrow a=b=c=1\)
Bài 1:
dự đoán dấu = sẽ là \(a^2=b^2=c^2=\dfrac{1}{2}\) nên cứ thế mà chém thôi .
Ta có: \(\left(a^2+1\right)\left(b^2+1\right)=\left(a^2+\dfrac{1}{2}\right)\left(\dfrac{1}{2}+b^2\right)+\dfrac{1}{2}\left(a^2+b^2\right)+\dfrac{3}{4}\)
Bunyakovsky:\(\left(a^2+\dfrac{1}{2}\right)\left(\dfrac{1}{2}+b^2\right)+\dfrac{1}{2}\left(a^2+b^2\right)+\dfrac{3}{4}\ge\dfrac{1}{2}\left(a+b\right)^2+\dfrac{1}{4}\left(a+b\right)^2+\dfrac{3}{4}=\dfrac{3}{4}\left[\left(a+b\right)^2+1\right]\)
\(VT=\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)\ge\dfrac{3}{4}\left[\left(a+b\right)^2+1\right]\left(1+c^2\right)\ge\dfrac{3}{4}\left(a+b+c\right)^2\)(đpcm)
Dấu = xảy ra khi \(a=b=c=\dfrac{1}{\sqrt{2}}\)
P/s: còn 1 cách khác nữa đó là khai triển sau đó xài schur . Chi tiết trong tệp BĐT schur .pdf