Cho a,b,c>0, chứng minh \(\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{a+b}\ge\frac{a+b+c}{2}\)
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Ta chứng minh BĐT sau với các số dương:
\(\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{4}{x+y}\)
Thật vậy, BĐT tương đương: \(\dfrac{x+y}{xy}\ge\dfrac{4}{x+y}\Leftrightarrow\left(x+y\right)^2\ge4xy\)
\(\Leftrightarrow x^2-2xy+y^2\ge0\Leftrightarrow\left(x-y\right)^2\ge0\) (luôn đúng)
Áp dụng:
\(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\) ; \(\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{4}{b+c}\) ; \(\dfrac{1}{c}+\dfrac{1}{a}\ge\dfrac{4}{c+a}\)
Cộng vế với vế:
\(2\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge\dfrac{4}{a+b}+\dfrac{4}{b+c}+\dfrac{4}{c+a}\)
\(\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{2}{a+b}+\dfrac{2}{b+c}+\dfrac{2}{c+a}\)
b.
Ta có:
\(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\Rightarrow\dfrac{3}{a}+\dfrac{3}{b}\ge\dfrac{12}{a+b}\) (1)
\(\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{4}{b+c}\Rightarrow\dfrac{2}{b}+\dfrac{2}{c}\ge\dfrac{8}{b+c}\) (2)
\(\dfrac{1}{c}+\dfrac{1}{a}\ge\dfrac{4}{c+a}\) (3)
Cộng vế với vế (1); (2) và (3):
\(\dfrac{4}{a}+\dfrac{5}{b}+\dfrac{3}{c}\ge4\left(\dfrac{3}{a+b}+\dfrac{2}{b+c}+\dfrac{1}{c+a}\right)\) (đpcm)
Dấu "=" xảy ra khi \(a=b=c\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Đặt: \(b+c=x;c+a=y;a+b=z\)
Có: \(x+y-z=b+c+c+a-a-b=2c\)
=> \(c=\frac{x+y-z}{2}\)
Tương tự ta cũng có:
\(a=\frac{y+z-x}{2};b=\frac{x+z-y}{2}\)
Có: \(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\)
=\(\frac{y+z-x}{2x}+\frac{x+z-y}{2y}+\frac{x+y-z}{2z}\)
\(=\frac{1}{2}\left(\frac{y}{x}+\frac{z}{x}-1+\frac{x}{y}+\frac{z}{y}-1+\frac{x}{z}+\frac{y}{z}-1\right)\)
\(=\frac{1}{2}\left[\left(\frac{y}{x}+\frac{x}{y}\right)+\left(\frac{z}{x}+\frac{x}{z}\right)+\left(\frac{z}{y}+\frac{y}{z}\right)-3\right]\) (1)
Áp dụng bđt cô si ta có:
\(\frac{y}{x}+\frac{x}{y}\ge2;\frac{z}{x}+\frac{x}{z}\ge2;\frac{z}{y}+\frac{y}{z}\ge2\)
=> \(\left(1\right)\ge\frac{1}{2}\left(2+2+2-3\right)=\frac{3}{2}\)
Vậy \(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge\frac{3}{2}\)
b) Có: \(\frac{a^2}{b+c}+\frac{b+c}{4}=\frac{\left(2a\right)^2+\left(b+c\right)^2}{4\left(b+c\right)}\) (1)
VÌ: \(\left[2a-\left(b+c\right)\right]^2\ge0\)
=> \(\left(2a\right)^2+\left(b+c\right)^2\ge4a\left(b+c\right)\)
=> \(\left(1\right)\ge\frac{4a\left(b+c\right)}{4\left(b+c\right)}=a\)
Hay: \(\frac{a^2}{b+c}+\frac{b+c}{4}\ge a\Rightarrow\frac{a^2}{b+c}\ge a-\frac{b+c}{4}\) (2)
Tương tự ta cũng có: \(\frac{b^2}{c+a}\ge b-\frac{c+a}{4}\) (3)
\(\frac{c^2}{a+b}\ge c-\frac{a+b}{4}\) (4)
Cộng vế với vế (2);(3);(4) ta có:
\(\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\ge a+b+c-\left(\frac{b+c+c+a+a+b}{4}\right)=\left(a+b+c\right)-\frac{a+b+c}{2}=\frac{a+b+c}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
Ta có:
\(\frac{a^2}{b^2+c^2}+\frac{b^2}{a^2+c^2}+\frac{c^2}{a^2+b^2}\geq \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\)
\(\Leftrightarrow \left(\frac{a^2}{b^2+c^2}-\frac{a}{b+c}\right)+\left(\frac{b^2}{a^2+c^2}-\frac{b}{a+c}\right)+\left(\frac{c^2}{a^2+b^2}-\frac{c}{a+b}\right)\geq 0\)
\(\Leftrightarrow \frac{ab(a-b)+ac(a-c)}{(b^2+c^2)(b+c)}+\frac{ba(b-a)+bc(b-c)}{(a^2+c^2)(a+c)}+\frac{ca(c-a)+cb(c-b)}{(a^2+b^2)(a+b)}\geq 0\)
\(\Leftrightarrow ab(a-b)\left(\frac{1}{(b^2+c^2)(b+c)}-\frac{1}{(a^2+c^2)(a+c)}\right)+bc(b-c)\left(\frac{1}{(a^2+c^2)(a+c)}-\frac{1}{(a^2+b^2)(a+b)}\right)+ca(c-a)\left(\frac{1}{(a^2+b^2)(a+b)}-\frac{1}{(b^2+c^2)(b+c)}\right)\geq 0\)
\(\Leftrightarrow ab(a-b).\frac{(a-b)(a^2+b^2+c^2+ab+bc+ac)}{(b^2+c^2)(b+c)(a^2+c^2)(a+c)}+bc(b-c).\frac{(b-c)(a^2+b^2+c^2+ab+bc+ac)}{(a^2+c^2)(a+c)(a^2+b^2)(a+b)}+ca(c-a).\frac{(c-a)(a^2+b^2+c^2+ab+bc+ac)}{(a^2+b^2)(a+b)(b^2+c^2)(b+c)}\geq 0\)
\(\Leftrightarrow (a^2+b^2+c^2+ab+bc+ac)\left[\frac{(a-b)^2}{(b^2+c^2)(b+c)(a^2+c^2)(a+c)}+\frac{(b-c)^2}{(a^2+c^2)(a+c)(a^2+b^2)(a+b)}+\frac{(c-a)^2}{(a^2+b^2)(a+b)(b^2+c^2)(b+c)}\right]\geq 0\)
(luôn đúng)
Ta có đpcm
Dấu "=" xảy ra khi $a=b=c$
![](https://rs.olm.vn/images/avt/0.png?1311)
Do vai trò của a,b,c là như nhau nên ta cò thể giả sử: \(a\ge b\ge c>0\)
Ta có:\(\frac{a^2}{b^2+c^2}-\frac{a}{b+c}=\frac{ab\left(a-b\right)+ac\left(a-c\right)}{\left(b+c\right)\left(b^2+c^2\right)}\)
CMTT: \(\frac{b^2}{c^2+a^2}-\frac{b}{c+a}=\frac{bc\left(b-c\right)-ab\left(a-b\right)}{\left(a+c\right)\left(a^2+c^2\right)}\)
\(\frac{c^2}{a^2+b^2}-\frac{c}{a+b}=\frac{-bc\left(b-c\right)-ac\left(a-c\right)}{\left(a+b\right)\left(a^2+b^2\right)}\)
Đặt \(A=\left(\frac{a^2}{b^2+c^2}+\frac{b^2}{c^2+a^2}+\frac{c^2}{a^2+b^2}\right)-\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)\)
\(\Rightarrow A=\left[\frac{ab\left(a-b\right)}{\left(b+c\right)\left(b^2+c^2\right)}-\frac{ab\left(a-b\right)}{\left(a+c\right)\left(a^2+c^2\right)}\right]\)+ \(\left[\frac{ac\left(a-c\right)}{\left(b+c\right)\left(b^2+c^2\right)}-\frac{ac\left(a-c\right)}{\left(a+b\right)\left(a^2+b^2\right)}\right]\)+ \(\left[\frac{bc\left(b-c\right)}{\left(a+c\right)\left(a^2+c^2\right)}-\frac{bc\left(b-c\right)}{\left(a+b\right)\left(a^2+b^2\right)}\right]\)
\(\Rightarrow A=ab\left(a-b\right)\left[\frac{1}{\left(b+c\right)\left(b^2+c^2\right)}-\frac{1}{\left(a+c\right)\left(a^2+c^2\right)}\right]^{\left(1\right)}\)+ ...
Do \(a\ge b\ge c>0\Rightarrow\left(1\right)>0.\)
CMTT \(\Rightarrow A>0.\Rightarrowđpcm\)
(Mình làm hơi tắt, mong bạn thông cảm. Cho 1 k nha.)
Tại sao (1) lại >0 hả bạn? Với lại đề mình cho đâu có đk a>=b>=c>0 đâu!
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2) Theo nguyên lí Dirichlet, trong ba số \(a^2-1;b^2-1;c^2-1\) có ít nhất hai số nằm cùng phía với 1.
Giả sử đó là a2 - 1 và b2 - 1. Khi đó \(\left(a^2-1\right)\left(b^2-1\right)\ge0\Leftrightarrow a^2b^2-a^2-b^2+1\ge0\)
\(\Rightarrow a^2b^2+3a^2+3b^2+9\ge4a^2+4b^2+8\)
\(\Rightarrow\left(a^2+3\right)\left(b^2+3\right)\ge4\left(a^2+b^2+2\right)\)
\(\Rightarrow\left(a^2+3\right)\left(b^2+3\right)\left(c^2+3\right)\ge4\left(a^2+b^2+1+1\right)\left(1+1+c^2+1\right)\) (2)
Mà \(4\left[\left(a^2+b^2+1+1\right)\left(1+1+c^2+1\right)\right]\ge4\left(a+b+c+1\right)^2\) (3)(Áp dụng Bunhicopxki và cái ngoặc vuông)
Từ (2) và (3) ta có đpcm.
Sai thì chịu
Xí quên bài 2 b:v
b) Không mất tính tổng quát, giả sử \(\left(a^2-\frac{1}{4}\right)\left(b^2-\frac{1}{4}\right)\ge0\)
Suy ra \(a^2b^2-\frac{1}{4}a^2-\frac{1}{4}b^2+\frac{1}{16}\ge0\)
\(\Rightarrow a^2b^2+a^2+b^2+1\ge\frac{5}{4}a^2+\frac{5}{4}b^2+\frac{15}{16}\)
Hay \(\left(a^2+1\right)\left(b^2+1\right)\ge\frac{5}{4}\left(a^2+b^2+\frac{3}{4}\right)\)
Suy ra \(\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)\ge\frac{5}{4}\left(a^2+b^2+\frac{1}{4}+\frac{1}{2}\right)\left(\frac{1}{4}+\frac{1}{4}+c^2+\frac{1}{2}\right)\)
\(\ge\frac{5}{4}\left(\frac{1}{2}a+\frac{1}{2}b+\frac{1}{2}c+\frac{1}{2}\right)^2=\frac{5}{16}\left(a+b+c+1\right)^2\) (Bunhiacopxki) (đpcm)
Đẳng thức xảy ra khi \(a=b=c=\frac{1}{2}\)
\(\frac{a^2}{b+c}+\frac{b+c}{4}\ge2\sqrt{\frac{a^2\left(b+c\right)}{4\left(b+c\right)}}=a\)
Tương tự: \(\frac{b^2}{a+c}+\frac{a+c}{4}\ge b\) ; \(\frac{c^2}{a+b}+\frac{a+b}{4}\ge c\)
Cộng vế với vế:
\(VT+\frac{a+b+c}{2}\ge a+b+c\Rightarrow VT\ge\frac{a+b+c}{2}\)
Dấu "=" xảy ra khi \(a=b=c\)