Do a,b,c dương
Để làm bài này bạn cần chứng minh BĐT sau\(\dfrac{x^2}{m}+\dfrac{y^2}{n}\ge\dfrac{\left(x+y\right)^2}{m+n}\)(m;n>0)
<=>(m+n)(nx2+my2)-mn(x+y)2\(\ge\)0
Mình làm tắt,rút gọn luôn
<=>n2x2-2mnxy+m2y2\(\ge\)0
<=>(nx-my)2\(\ge\)0
=>BĐT trên được chứng minh và dấu bằng xảy ra khi nx=my
Mở rộng cho 3 số \(\dfrac{x^2}{m}+\dfrac{y^2}{n}+\dfrac{z^2}{p}\ge\dfrac{\left(x+y+z\right)^2}{m+n+p}\)
Áp dụng BĐT trên ta được:
\(\dfrac{a^2}{b+c}+\dfrac{b^2}{a+c}+\dfrac{c^2}{a+b}\ge\dfrac{\left(a+b+c\right)^2}{2\left(a+b+c\right)}=\dfrac{a+b+c}{2}\)
Dấu = xảy ra khi a=b=c
Mk giải dài dòng một chút
\(\dfrac{a^2}{b+c}+\dfrac{b^2}{a+c}+\dfrac{c^2}{a+b}>=\dfrac{a+b+c}{2}\)
Chứng minh bài toán phụ: \(\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{b+a}>=\dfrac{3}{2}\)
Ta có BĐT: \(\left(a+b+c\right).\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)>=9\)
Áp dụng bài toán phụ trên:
=> \(\left(a+b+b+c+c+a\right)\left(\dfrac{1}{a+b}+\dfrac{1}{a+c}+\dfrac{1}{b+c}\right)\)>=9
<=> \(\dfrac{2a}{a+b}+\dfrac{2a}{a+c}+\dfrac{2a}{b+c}+\dfrac{2b}{a+b}+\dfrac{2b}{a+c}+\dfrac{2b}{b+c}+\dfrac{2c}{a+b}+\dfrac{2c}{a+c}+\dfrac{2c}{b+c}>=9\)
<=> \(\left(\dfrac{2a}{a+b}+\dfrac{2b}{a+b}+\dfrac{2c}{a+b}\right)\)+\(\left(\dfrac{2a}{a+c}+\dfrac{2b}{a+c}+\dfrac{2c}{a+c}\right)\)+\(\left(\dfrac{2a}{b+c}+\dfrac{2b}{b+c}+\dfrac{2c}{b+c}\right)\)>=9
<=> \(\dfrac{2c}{a+b}+2+\dfrac{2b}{a+c}+2+\dfrac{2a}{b+c}+2>=9\)
<=>\(2.\left(\dfrac{c}{a+b}+\dfrac{b}{a+c}+\dfrac{a}{b+c}\right)>=3\)
<=> \(\dfrac{c}{a+b}+\dfrac{b}{a+c}+\dfrac{a}{b+c}>=\dfrac{3}{2}\) (1)
+) \(\dfrac{a^2}{b+c}+a+\dfrac{b^2}{a+c}+b+\dfrac{c^2}{a+b}+c\)
= \(\dfrac{a^2+ab+ac}{b+c}\)+ \(\dfrac{b^2+ba+bc}{a+c}\)+\(\dfrac{c^2+ac+bc}{b+a}\)
= \(\dfrac{a\left(a+b+c\right)}{b+c}+\dfrac{b\left(a+b+c\right)}{a+c}+\dfrac{c\left(a+b+c\right)}{b+a}\)
= \(\left(a+b+c\right)\left(\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{b+a}\right)\)(2)
Từ (1) và(2) => \(\left(a+b+c\right)\left(\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b}\right)>=\)\(\dfrac{3}{2}\left(a+b+c\right)\) (3)
+) \(\dfrac{a+b+c}{2}+a+b+c\)
= \(\dfrac{a+b+c+2a+2b+2c}{2}\)
= \(\dfrac{3a+3b+3c}{2}\)
= \(\dfrac{3}{2}\left(a+b+c\right)\) (4)
Từ (3) và (4) =>
\(\dfrac{a^2}{b+c}+\dfrac{b^2}{a+c}+\dfrac{c^2}{a+b}+a+b+c\)>=\(\dfrac{a+b+c}{2}+a+b+c\)
=> \(\dfrac{a^2}{b+c}+\dfrac{b^2}{a+c}+\dfrac{c^2}{a+b}>=\dfrac{a+b+c}{2}\) => đpcm
Mk làm hơi dài dòng. Bạn chịu khó đọc nha
áp dụng bất đẳng thức cauchy schwarz dạng engel với các số k âm, ta có
\(\dfrac{a^2}{b+c}+\dfrac{b^2}{a+c}+\dfrac{c^2}{a+b}\ge\dfrac{\left(a+b+c\right)^2}{2\left(a+b+c\right)}=\dfrac{a+b+c}{2}\)
\(\left(a+b+c\right)^2=\left(\dfrac{a}{\sqrt{b+c}}.\sqrt{b+c}+\dfrac{b}{\sqrt{a+c}}.\sqrt{a+c}+\dfrac{c}{\sqrt{a+b}}.\sqrt{a+b}\right)^2\le\left(\dfrac{a^2}{b+c}+\dfrac{b^2}{a+c}+\dfrac{c^2}{a+b}\right).2\left(a+b+c\right)\)
\(\Rightarrow\left(\dfrac{a^2}{b+c}+\dfrac{b^2}{a+c}+\dfrac{c^2}{a+b}\right)\ge\dfrac{a+b+c}{2}\)
dấu "=" xảy ra khi a=b=c.
Áp dụng BĐT AM-GM cho 2 số ta có
\(\dfrac{a^2}{b+c}+\dfrac{b+c}{4}\ge a\)
\(\dfrac{b^2}{a+c}+\dfrac{a+c}{4}\ge b\)
\(\dfrac{c^2}{a+b}+\dfrac{a+b}{4}\ge c\)
cộng các vế của BĐT trên ta có đpcm
đẳng thức xảy ra khi a=b=c
