chứng minh rằng \(\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x-4\right)\ge-1\)
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Xét hiệu :
\(\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x-4\right)-\left(-1\right)=\left(x^2-5x+4\right)\left(x^2-5x+6\right)+1.\)
Đặt \(x^2-5x+=y.\) Biểu thức trên bằng \(\left(y-1\right)\left(y+1\right)+1=y^2\ge0\)
Vậy \(\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x-4\right)\ge-1\)
Em thử ạ!Em không chắc đâu.Hơi quá sức em rồi
Ta có: \(VT=\Sigma\frac{x^3}{z+y+yz+1}=\Sigma\frac{x^3}{z+y+\frac{1}{x}+1}\)
\(=\Sigma\frac{x^4}{xz+xy+1+x}=\frac{x^4}{xy+xz+x+1}+\frac{y^4}{yz+xy+y+1}+\frac{z^4}{zx+yz+z+1}\)
Áp dụng BĐT Cauchy-Schwarz dạng Engel,suy ra:
\(VT\ge\frac{\left(x^2+y^2+z^2\right)^2}{\left(x+y+z\right)+2\left(xy+yz+zx\right)+3}\)
\(\ge\frac{\left(\frac{1}{3}\left(x+y+z\right)^2\right)^2}{\left(x+y+z\right)+\frac{2}{3}\left(x+y+z\right)^2+3}\) (áp dụng BĐT \(a^2+b^2+c^2\ge\frac{\left(a+b+c\right)^2}{3};ab+bc+ca\le\frac{\left(a+b+c\right)^2}{3}\))
Đặt \(t=x+y+z\ge3\sqrt{xyz}=3\) Dấu "=" xảy ra khi x = y = z
Ta cần chứng minh: \(\frac{\frac{t^4}{9}}{\frac{2}{3}t^2+t+3}\ge\frac{3}{4}\Leftrightarrow\frac{t^4}{9\left(\frac{2}{3}t^2+t+3\right)}=\frac{t^4}{6t^2+9t+27}\ge\frac{3}{4}\)(\(t\ge3\))
Thật vậy,BĐT tương đương với: \(4t^4\ge18t^2+27t+81\)
\(\Leftrightarrow3t^4-18t^2-27t+t^4-81\ge0\)
Ta có: \(VT\ge3t^4-18t^2-27t+3^4-81\)
\(=3t^4-18t^2-27t\).Cần chứng minh\(3t^4-18t^2-27t\ge0\Leftrightarrow3t^4\ge18t^2+27t\)
Thật vậy,chia hai vế cho \(t\ge3\),ta cần chứng minh \(3t^3\ge18t+27\Leftrightarrow3t^3-18t-27\ge0\)
\(\Leftrightarrow3\left(t^3-27\right)-18\left(t-3\right)\ge0\)
\(\Leftrightarrow\left(t-3\right)\left(3t^2+9t+27\right)-18\left(t-3\right)\ge0\)
\(\Leftrightarrow\left(t-3\right)\left(3t^2+9t+9\right)\ge0\)
BĐT hiển nhiên đúng,do \(t\ge3\) và \(3t^2+9t+9=3\left(t+\frac{3}{2}\right)^2+\frac{9}{4}\ge\frac{9}{4}>0\)
Dấu "=" xảy ra khi t = 3 tức là \(\hept{\begin{cases}x=y=z\\xyz=1\end{cases}}\Leftrightarrow x=y=z=1\)
Chứng minh hoàn tất
Em sửa chút cho bài làm ngắn gọn hơn.
Khúc chứng minh: \(4t^4\ge18t^2+27t+81\)
\(\Leftrightarrow4t^4-18t^2-27t-81\ge0\)
\(\Leftrightarrow\left(t-3\right)\left(4t^3+12t^2+18t+27\right)\ge0\)
BĐT hiển nhiên đúng do \(t\ge3\Rightarrow\hept{\begin{cases}t-3\ge0\\4t^3+12t^2+18t+27>0\end{cases}}\)
Còn khúc sau y chang :P Lúc làm rối quá nên không nghĩ ra ạ!
\(\frac{x^3}{\left(1+y\right)\left(1+z\right)}+\frac{y^3}{\left(1+z\right)\left(1+x\right)}+\frac{z^3}{\left(1+x\right)\left(1+y\right)}\ge\frac{3}{4}\)
\(=\frac{x^3}{1+z+y+yz}+\frac{y^3}{1+x+z+xz}+\frac{z^3}{1+y+x+xy}\)
\(=\frac{x^3}{1+x+y+2y}\ge\frac{x}{2}\Rightarrow TổngBPT\ge\frac{x}{2}+\frac{y}{2}+\frac{z}{2}\ge\frac{2}{3}\left(đpcm\right)\)
(Không chắc à nha)
Ta có : \(\frac{x^3}{\left(1+y\right)\left(1+z\right)}+\frac{1+y}{8}+\frac{1+z}{8}\ge\frac{3x}{4}\)
\(\Rightarrow\frac{x^3}{\left(1+y\right)\left(1+z\right)}\ge\frac{6x-y-z-2}{8}\left(1\right)\)
Tương tự ta có : \(\hept{\begin{cases}\frac{y^3}{\left(1+z\right)\left(1+x\right)}\ge\frac{6y-z-x-2}{8}\left(2\right)\\\frac{z^3}{\left(1+x\right)\left(1+y\right)}\ge\frac{6z-x-y-2}{8}\left(3\right)\end{cases}}\)
Từ (1) , (2) và (3)
\(\Rightarrow\frac{x^3}{\left(1+y\right)\left(1+z\right)}+\frac{y^3}{\left(1+z\right)\left(1+x\right)}+\frac{z^3}{\left(1+x\right)\left(1+y\right)}\)
\(\ge\frac{6x-y-z-2}{8}+\frac{6y-z-x-2}{8}+\frac{6z-x-y-2}{8}\)
\(=\frac{1}{2}\left(x+y+z\right)-\frac{3}{4}\ge\frac{3}{2}-\frac{3}{4}=\frac{3}{4}\)
Chúc bạn học tốt !!!
Ta có : \(\frac{x^3}{\left(1+y\right)\left(1+z\right)}+\frac{1+y}{8}+\frac{1+z}{8}\ge\frac{3x}{4}\)
\(\Rightarrow\frac{x^3}{\left(1+y\right)\left(1+z\right)}\ge\frac{6x-y-z-2}{8}\left(1\right)\)
Tương tự ta có : \(\hept{\begin{cases}\frac{y^3}{\left(1+z\right)\left(1+x\right)}\ge\frac{6y-z-x-2}{8}\left(2\right)\\\frac{z^3}{\left(1+x\right)\left(1+y\right)}\ge\frac{6z-x-y-2}{8}\left(3\right)\end{cases}}\)
Từ (1) , (2) , (3)
\(\Rightarrow\frac{x^3}{\left(1+y\right)\left(1+z\right)}+\frac{y^3}{\left(1+z\right)\left(1+x\right)}+\frac{z^3}{\left(1+x\right)\left(1+y\right)}\)
\(\ge\frac{6x-y-z-2}{8}+\frac{6y-z-x-2}{8}+\frac{6z-x-y-2}{8}\)
\(=\frac{1}{2}\left(x+y+z\right)-\frac{3}{4}\ge\frac{3}{2}-\frac{3}{4}=\frac{3}{4}\)
Chúc bạn học tốt !!!
Áp dụng bđt AM-GM ta có:
\(\hept{\begin{cases}\frac{x^3}{\left(1+y\right)\left(1+z\right)}+\frac{1+y}{8}+\frac{1+z}{8}\ge3\sqrt[3]{\frac{x^3}{\left(1+y\right)\left(1+z\right)}.\frac{1+y}{8}.\frac{1+z}{8}}=\frac{3x}{4}\left(1\right)\\\frac{y^3}{\left(1+z\right)\left(1+x\right)}+\frac{1+z}{8}+\frac{1+x}{8}\ge3\sqrt[3]{\frac{y^3}{\left(1+z\right)\left(1+x\right)}.\frac{1+z}{8}.\frac{1+x}{8}}=\frac{3y}{4}\left(2\right)\\\frac{z^3}{\left(1+x\right)\left(1+y\right)}+\frac{1+x}{8}+\frac{1+y}{8}\ge3\sqrt[3]{\frac{z^3}{\left(1+x\right)\left(1+y\right)}.\frac{1+x}{8}.\frac{1+y}{8}}=\frac{3z}{4}\left(3\right)\end{cases}}\)
Lấy \(\left(1\right)+\left(2\right)+\left(3\right)\)ta được:
\(P+\frac{3+x+y+z}{4}\ge\frac{3\left(x+y+z\right)}{4}\)
\(\Leftrightarrow P\ge\frac{3\left(x+y+z\right)}{4}-\frac{3+x+y+z}{4}\)
\(\Leftrightarrow P\ge\frac{2\left(x+y+z\right)-3}{4}\left(1\right)\)
Áp dụng bdt AM-GM ta có:
\(x+y+z\ge3\sqrt[3]{xyz}=3\)Thay vào (1) ta được:
\(P\ge\frac{2.3-3}{4}\)
\(\Rightarrow P\ge\frac{3}{4}\)Dấu"="xảy ra \(\Leftrightarrow x=y=z\)
Ta có: \(\dfrac{x^3}{\left(1+y\right)\left(1+z\right)}+\dfrac{1+y}{8}+\dfrac{1+z}{8}\ge\dfrac{3x}{4}\)
\(\Rightarrow\dfrac{x^3}{\left(1+y\right)\left(1+z\right)}\ge\dfrac{6x-y-z-2}{8}\left(1\right)\)
Tương tự ta có: \(\left\{{}\begin{matrix}\dfrac{y^3}{\left(1+z\right)\left(1+x\right)}\ge\dfrac{6y-z-x-2}{8}\left(2\right)\\\dfrac{z^3}{\left(1+x\right)\left(1+y\right)}\ge\dfrac{6z-x-y-2}{8}\left(3\right)\end{matrix}\right.\)
Từ (1), (2), (3)
\(\Rightarrow\dfrac{x^3}{\left(1+y\right)\left(1+z\right)}+\dfrac{y^3}{\left(1+z\right)\left(1+x\right)}+\dfrac{z^3}{\left(1+x\right)\left(1+y\right)}\ge\dfrac{6x-y-z-2}{8}+\dfrac{6y-z-x-2}{8}+\dfrac{6z-x-y-2}{8}\)
\(=\dfrac{1}{2}\left(x+y+z\right)-\dfrac{3}{4}\ge\dfrac{3}{2}-\dfrac{3}{4}=\dfrac{3}{4}\)
Bài dễ mừ, có phải Croatia thật ko vậy :)) (viết đề bị nhầm, là x,y,z dương chứ :))
Áp dụng Cauchy-Schwarz dạng cộng mẫu số:
\(\frac{x^2}{\left(x+y\right)\left(x+z\right)}+\frac{y^2}{\left(y+z\right)\left(y+x\right)}+\frac{z^2}{\left(z+x\right)\left(z+y\right)}\ge\)
\(\frac{\left(x+y+z\right)^2}{\left(x+y\right)\left(x+z\right)+\left(y+z\right)\left(y+x\right)+\left(z+x\right)\left(z+y\right)}=\frac{\left(x+y+z\right)^2}{x^2+y^2+z^2+3\left(xy+yz+zx\right)}\)
\(=\frac{\left(x+y+z\right)^2}{\left(x+y+z\right)^2+\left(xy+yz+zx\right)}\)
Xét \(xy+yz+zx\le\frac{\left(x+y+z\right)^2}{3}\Rightarrow\frac{\left(x+y+z\right)^2}{\left(x+y+z\right)^2+\left(xy+yz+zx\right)}\ge\frac{\left(x+y+z\right)^2}{\left(x+y+z\right)^2+\frac{\left(x+y+z\right)^2}{3}}\)
\(=\frac{\left(x+y+z\right)^2}{\frac{4}{3}\left(x+y+z\right)^2}=\frac{3}{4}\)
Dấu bằng xảy ra khi và chỉ khi x=y=z, Xong! :))
\(A=x^2-16-6x-2x^2+x^2+6x+9=-7\\ B=\left(x^2+4\right)\left(x^2-4\right)-x^4+9\\ B=x^4-16-x^4+9=-7\)
a) \(A=\left(x+4\right)\left(x-4\right)-2x\left(3+x\right)+\left(x+3\right)^2\)
\(=x^2-16-2x^2-6x+x^2+6x+9=-7\)
b) \(B=\left(x^2+4\right)\left(x+2\right)\left(x-2\right)-\left(x^2+3\right)\left(x^2-3\right)\)
\(=\left(x^2+4\right)\left(x^2-4\right)-\left(x^4-9\right)\)
\(=x^4-16-x^4+9=-7\)
Xét hiệu : \(\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x-4\right)-\left(-1\right)=\left(x^2-5x+4\right)\left(x^2-5x+6\right)+1\)
Đặt \(x^2-5x+5=y\). Biểu thức trên bằng :\(\left(y-1\right)\left(y+1\right)+1=y^2\ge0\)
Vậy \(\text{ ( x − 1 ) ( x − 2 ) ( x − 3 ) ( x − 4 ) ≥ − 1}\)