Supppose that \(\left|ax^2+bx+c\right|\ge\left|x^2-1\right|\) for all real numbers x. Prove that \(\left|b^2-4ac\right|\ge4\)
Let x, y, z be positive real numbers such that xy + yz + zx + xyz = 4 . Prove that :
\(3\left(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}+\frac{1}{\sqrt{z}}\right)^2\ge\left(x+2\right)\left(y+2\right)\left(z+2\right)\)
Đặt \(x=\frac{2a}{b+c};y=\frac{2b}{c+a};z=\frac{2c}{a+b}\) Thì bài toán thành chứng minh
\(3\left(\sqrt{\frac{a+b}{2c}}+\sqrt{\frac{b+c}{2a}}+\sqrt{\frac{c+a}{2b}}\right)^2\ge\frac{8\left(a+b+c\right)^3}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)
Áp dụng holder ta có:
\(\left(\sqrt{\frac{a+b}{2c}}+\sqrt{\frac{b+c}{2a}}+\sqrt{\frac{c+a}{2b}}\right)^2\left(2c\left(a+b\right)^2+2a\left(b+c\right)^2+2b\left(c+a\right)^2\right)\)
\(\ge\left[\left(a+b\right)+\left(b+c\right)+\left(c+a\right)\right]^3=8\left(a+b+c\right)^3\)
\(\Rightarrow VT\ge3.\frac{8\left(a+b+c\right)^3}{2a\left(b+c\right)^2+2b\left(c+a\right)^2+2c\left(a+b\right)^2}\)
Từ đây ta cần chứng minh:
\(3.\frac{8\left(a+b+c\right)^3}{2a\left(b+c\right)^2+2b\left(c+a\right)^2+2c\left(a+b\right)^2}\ge\frac{8\left(a+b+c\right)^3}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)
\(\Leftrightarrow2a\left(b+c\right)^2+2b\left(c+a\right)^2+2c\left(a+b\right)^2\le3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
\(\Leftrightarrow a\left(b-c\right)^2+b\left(c-a\right)^2+c\left(a-b\right)^2\ge0\)( đúng )
Vậy có ĐPCM
Giả sử \(\left|ã^2+bx+c\right|\ge\left|x^2-1\right|\) đúng với mọi số thực x. Chứng minh rằng \(\left|b^2-4ac\right|\ge4\)
a=1;b=2;c=3;x=8 tin to di bai nay to hoc roi\
lamf tis baats ddawngr thwcs :))
Let a,b,c be positive real numbers. Prove that : \(\frac{1}{a\left(b+1\right)}+\frac{1}{b\left(c+1\right)}+\frac{1}{c\left(a+1\right)}\ge\frac{3}{\sqrt[3]{abc}\left(1+\sqrt[3]{abc}\right)}\)
Đặt \(\left(a;b;c\right)=\left(\frac{x}{y}k;\frac{y}{z}k;\frac{z}{x}k\right)\) \(k\inℝ^+\)
Bất đẳng thức cần chứng minh tương đương:
\(\frac{1}{\frac{x}{y}k\left(\frac{y}{z}k+1\right)}+\frac{1}{\frac{y}{z}k\left(\frac{z}{x}k+1\right)}+\frac{1}{\frac{z}{x}k\left(\frac{x}{y}k+1\right)}\ge\frac{3}{\sqrt[3]{\frac{x}{y}k\cdot\frac{y}{z}k\cdot\frac{z}{x}k}\left(1+\sqrt[3]{\frac{x}{y}k\cdot\frac{y}{z}k\cdot\frac{z}{x}k}\right)}\)
\(\Leftrightarrow\frac{yz}{xk\left(yk+z\right)}+\frac{zx}{yk\left(zk+x\right)}+\frac{xy}{zk\left(xk+y\right)}\ge\frac{3}{k\left(1+k\right)}\) (D)
Ta có: \(\frac{yz}{xk\left(yk+z\right)}+\frac{zx}{yk\left(zk+x\right)}+\frac{xy}{zk\left(xk+y\right)}\)
\(=\frac{\left(yz\right)^2}{xyzk\left(yk+z\right)}+\frac{\left(zx\right)^2}{xyzk\left(zk+x\right)}+\frac{\left(xy\right)^2}{xyzk\left(xk+y\right)}\)
\(\ge\frac{\left(xy+yz+zx\right)^2}{xyzk\left(xk+yk+zk+x+y+z\right)}\) (Bất đẳng thức Bunyakovsky dạng phân thức)
\(\ge\frac{3\left(xyz^2+xy^2z+x^2yz\right)}{xyzk\left(x+y+z\right)\left(k+1\right)}=\frac{3xyz\left(x+y+z\right)}{xyzk\left(x+y+z\right)\left(k+1\right)}=\frac{3}{k\left(k+1\right)}\)
=> BĐT (D) đúng => đpcm
Dấu "=" xảy ra khi: \(a=b=c\)
Let a;b;c be positive real numbers such that \(a^2+b^2+c^2=\frac{1}{3}\). Prove that :
\(4\left(a+b+c\right)+\frac{2}{3}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge10\)
We have:\(\hept{\begin{cases}a^2+b^2+c^2=\frac{1}{3}\\a,b,c>0\end{cases}\Rightarrow0< a,b,c< \frac{1}{\sqrt{3}}}\)
We prove to:
\(4x+\frac{2}{3x}\ge-3x^2+\frac{11}{3}\) with \(0< x< \frac{1}{\sqrt{3}}\)
\(\Leftrightarrow4x+\frac{2}{3x}+3x^2-\frac{11}{3}\ge0\)
\(\Leftrightarrow9x^3+12x^2-11x+2\ge0\)
\(\Leftrightarrow\left(3x+1\right)^2\left(x+2\right)\ge0\) Always true to all \(0< x< \frac{1}{\sqrt{3}}\)
\(\Rightarrow VT\ge-3a^2+\frac{11}{3}-3b^2+\frac{11}{3}-3c^2+\frac{11}{3}\)
\(=-3\left(a^2+b^2+c^2\right)+11=-3.\frac{1}{3}+11=10\) \(\left(đpcm\right)\)
Đặt biểu thức trên là \(A\)
Ta có : \(A=\left(4a+\frac{2}{3a}\right)+\left(4b+\frac{2}{3b}\right)+\left(4c+\frac{2}{3c}\right)\)
Cần chứng minh \(4a+\frac{2}{3a}\ge-3a^2+\frac{11}{3}\) (*)
Thật vậy \(BĐT\Leftrightarrow4a+\frac{2}{3a}+3a^2-\frac{11}{3}\ge0\)
\(\Leftrightarrow\frac{12a^2+2+9a^3-11a}{3a}\ge0\Leftrightarrow\frac{\left(a+2\right)\left(3a-1\right)^2}{3a}\ge0\) (luôn đúng)
Tương tự : \(4b+\frac{2}{3b}\ge-3b^2+\frac{11}{3}\) và \(4c+\frac{2}{3c}\ge-3c^2+\frac{11}{3}\)
Cộng các bất dẳng thức vừa CM đc ta có :
\(A\ge-3\left(a^2+b^2+c^2\right)+\frac{11}{3}.3=-3.\frac{1}{3}+11=10\)
Dấu "=" xảy ra \(\Leftrightarrow a=b=c=\frac{1}{3}\)
Let \(a,b,c,k\) be positive real numbers such that \(k\left(ab+bc+ca\right)+2abc\le k^3\) . Prove that:
\(\left(1\right)k\left(a+b+c\right)\ge2\left(ab+bc+ca\right)\)
\(\left(2\right)k\left(a^3+b^3+c^3\right)\ge2\left(a^2b^2+b^2c^2+c^2a^2\right)\)
\(\left(3\right)k\left(a^{2n-1}+b^{2n-1}+c^{2n-1}\right)\ge2\left(a^nb^n+b^nc^n+c^na^n\right)\) \(\left(n\ge0;n\in R\right)\)
mày bị điên rồi hả câu hỏi thế này làm gì có người giải được
Find the values of a,b and c such that
\(\left(ax^2+bx+c\right)\left(x-1\right)=-5x^3+4x^2+3x-2\).
Answer: The values of a,b and c are ......... , respectively.
(used " ; " between the numbers)
mình ko bít tiếng anh bn dịch hộ mình đi
1.tìm a,b để:
a)\(x^3+ax+bx+6⋮\left(x-1\right)\)
b)\(x^4+ax^3+bx^2+5x+1⋮\left(x+1\right)^2\)
c)\(^{x^4+3x^3+ax^2+bx+5⋮\left(x-2\right)^2}\)
d)\(x^4+10x^3+ax^2+bx+7⋮\left(x+2\right)^2\)
e)\(x^4+ax^3+5x^2+bx+1⋮x-1\)
2.Cho a+b+c=0.tính\(\left(a+b+c\right)^3+\left(b+a-c\right)^3+\left(c+a-b\right)^3\)
bài 2:
\(A=\left(a+b+c\right)^3+\left(b+a-c\right)^3+\left(c+a-b\right)^3\)
\(=\left(c+b+a-2c\right)^3+\left(c+a+b-2b\right)^3\)
\(=\left(-2c\right)^3+\left(-2b\right)^3=-8\left(b+c\right)\)
sao nữa nhỉ :v
cho tam thức \(f\left(x\right)=ax^2+bx+c\left(a\ne0\right)\), \(\Delta=b^2-4ac\). ta có f(x)>0 với mọi x thuộc r khi và chỉ khi nào
f(x)>0 với mọi x khi và chỉ khi: \(\left\{{}\begin{matrix}\text{Δ}< 0\\a>0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}b^2-4ac< 0\\a>0\end{matrix}\right.\)
Cho \(\left\{{}\begin{matrix}a\ne0\\b^2-4ac< 2b-1\end{matrix}\right.\). Chứng minh hệ sau vô nghiệm:
\(\left\{{}\begin{matrix}ax^2+bx+c=y\\ay^2+by+c=z\\az^2+bz+c=x\end{matrix}\right.\)