Cho a2 - a + 2b + 4b2 - 4ab \(\le\)0. CM: 0 \(\le\)a - 2b \(\le\)1
Với a<2b<0, rút gọn \(\dfrac{1}{a-2b}\)√b2(a2-4ab+4b2)
\(\dfrac{1}{a-2b}.\sqrt{b^2\left(a^2-4ab+4b^2\right)}=\dfrac{1}{a-2b}.b.\left|a-2b\right|=\dfrac{1}{a-2b}.b.\left(2b-a\right)=-b\)
\(\dfrac{1}{a-2b}\cdot\sqrt{b^2\cdot\left(a^2-4ab+b^2\right)}\)
\(=\dfrac{1\cdot\left(a-2b\right)}{a-2b}\cdot b\)
=b
Cho \(0\le a\le b\le1\), chứng minh \(0\le ab^2-a^2b\le\frac{1}{4}\)
Cho a,b,c thỏa mãn a + b + c = 0 và -1 \(\le\)a, b, c \(\le\)1 .Tìm giá trị lớn nhất của biểu thức P = \(a^2+2b^2+c^2\)
cho 0< a < b ≤ 2 và 2ab ≤ 2b+a . Cmr: \(a^2+b^2\le5\)
Cho \(0\le a,b,c\le1\). Chứng minh \(a^2+b^2+c^2\le a^2b+b^2c+c^2a+1\)
Ta có \(a\left(1-a\right)\left(1-b\right)\ge0\)
\(\Leftrightarrow a^2b\ge a^2+ab-a\)
Tương tự \(b^2c\ge b^2+bc-b;c^2a\ge c^2+ca-a\)
\(\Rightarrow a^2b+b^2c+c^2a+1\ge a^2+b^2+c^2+ab+bc+ca-a-b-c+1\)\(=a^2+b^2+c^2+\left(1-a\right)\left(1-b\right)\left(1-c\right)+abc\ge a^2+b^2+c^2\)
Hay \(a^2+b^2+c^2\le a^2b+b^2c+c^2a+1\)
cho a,b,c thỏa mãn 0 ≤ a,b,c ≤ 1. Cmr: \(a^2+b^2+c^2\le1+a^2b+b^2c+c^2a\)
Lời giải:
Vì $a,b,c\in [0;1]$ nên: \(a(a-1)(b-1)\geq 0\)
\(\Leftrightarrow a(ab-a-b+1)\geq 0\)
\(\Leftrightarrow a^2b\geq a^2+ab-a\)
Tương tự với \(b^2c; c^2a\) suy ra:
\(a^2b+b^2c+c^2a+1\geq a^2+b^2+c^2+ab+bc+ac+1-a-b-c(1)\)
Lại có:
\((a-1)(b-1)(c-1)\leq 0\)
\(\Leftrightarrow (ab-a-b+1)(c-1)\leq 0\)
\(\Leftrightarrow abc-(ab+bc+ac)+a+b+c-1\leq 0\)
\(\Leftrightarrow ab+bc+ac+1\geq a+b+c+abc\geq a+b+c(2)\) do $abc\geq 0$
Từ \((1);(2)\Rightarrow a^2b+b^2c+c^2a+1\geq a^2+b^2+c^2\) (đpcm)
Cho \(0\le a\le b\le c\). CMR: \(\frac{2a^2}{b+c}+\frac{2b^2}{c+a}+\frac{2c^2}{a+b}\le\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\)
Theo bài ra ta có \(0\le a\le b\le c\) nên b\(+\)c \(\ge\)2b
Do đó suy ra \(\frac{2a^2}{b+c}\le\frac{2a^2}{2b}\)suy ra \(\frac{2a^2}{b+c}\le\frac{a^2}{b}\)
Chưng minh tương tự có \(\frac{2b^2}{c+a}\le\frac{b^2}{c}\)và \(\frac{2c^2}{a+b}\le\frac{c^2}{a}\)
Cộng vế với vế của các bđt cùng chiều trên ta sẽ suy ra điều phải chứng minh
#nga
Sai rồi nếu như theo cách chứng minh của bạn thì ta có: a + c \(\ge2c\)cái này vô lý.
Cho a,b,c > 0.CMR:
\(\frac{ab}{a^2+3b^2+4ab+5bc+3ac}+\frac{bc}{2a^2+b^2+3c^2+3ab+4bc+5ac}+\frac{ac}{3a^2+2b^2+c^2+5ab+3bc+4ac}\le\frac{1}{6}\)
a, a,b,c>0. CMR:\(\dfrac{ab}{a+b+2c}+\dfrac{bc}{b+c+2a}+\dfrac{ac}{a+c+2b}\le\dfrac{a+b+c}{4}\)
b, a,b,c>0. CMR:\(\dfrac{ab}{a+3b+2c}+\dfrac{bc}{b+3c+2a}+\dfrac{ac}{c+3a+2b}\le\dfrac{a+b+c}{6}\)
a.
\(\sum\dfrac{ab}{a+c+b+c}\le\dfrac{1}{4}\sum\left(\dfrac{ab}{a+c}+\dfrac{ab}{b+c}\right)=\dfrac{a+b+c}{4}\)
2.
\(\dfrac{ab}{a+3b+2c}=\dfrac{ab}{a+b+2c+2b}\le\dfrac{ab}{9}\left(\dfrac{4}{a+b+2c}+\dfrac{1}{2b}\right)=4.\dfrac{ab}{a+b+2c}+\dfrac{a}{18}\)
Quay lại câu a
\(b,\dfrac{ab}{a+3b+2c}=\left(\dfrac{1}{9}ab\right)\cdot\dfrac{9}{\left(a+c\right)+\left(b+c\right)+2b}\le\left(\dfrac{1}{9}ab\right)\cdot\left(\dfrac{1}{a+c}+\dfrac{1}{b+c}+\dfrac{1}{2b}\right)=\dfrac{1}{9}\cdot\left(\dfrac{ab}{a+b}+\dfrac{ab}{b+c}+\dfrac{a}{2}\right)\)
Cmtt: \(\dfrac{bc}{b+3c+2a}\le\dfrac{1}{9}\cdot\left(\dfrac{bc}{a+b}+\dfrac{bc}{a+b}+\dfrac{b}{2}\right);\dfrac{ca}{c+3a+2b}\le\dfrac{1}{9}\cdot\left(\dfrac{ca}{b+c}+\dfrac{ca}{a+b}+\dfrac{c}{2}\right)\)
\(\Rightarrow VT\le\dfrac{1}{9}\left(\dfrac{bc+ca}{a+b}+\dfrac{ab+ac}{b+c}+\dfrac{ab+bc}{a+c}+\dfrac{a+b+c}{2}\right)\\ \le\dfrac{1}{9}\left(a+b+c+\dfrac{a+b+c}{2}\right)=\dfrac{1}{9}\cdot\dfrac{3}{2}\left(a+b+c\right)=\dfrac{a+b+c}{6}\)
Dấu $"="$ khi $a=b=c$