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Băng Băng 2k6, Vũ Minh Tuấn, Nguyễn Việt Lâm, HISINOMA KINIMADO, Akai Haruma, Inosuke Hashibira,
Nguyễn Thị Ngọc Thơ, @tth_new
help me! cần gấp lắm ạ!
thanks nhiều!
Đặt cái ban đầu là A
Dầu tiên ta có
\(\text{(3a+c)(a+2b+c)+(3b+d)(b+2c+d)+(3c+a)(c+2d+a)+(3d+b)(d+2a+b)}\)
\(=4\left(a+b+c+d\right)^2\)
Ta có: \(\frac{a-b}{a+2b+c}+\frac{1}{2}=\frac{1}{2}.\frac{3a+c}{a+2b+c}=\frac{1}{2}.\frac{\left(3a+c\right)^2}{\left(3a+c\right)\left(a+2b+c\right)}\)
Tương tự ta có
\(\frac{b-c}{b+2c+d}+\frac{1}{2}=\frac{1}{2}.\frac{\left(3b+d\right)^2}{\left(3b+d\right)\left(b+2c+d\right)}\)
\(\frac{c-d}{c+2d+a}+\frac{1}{2}=\frac{1}{2}.\frac{\left(3c+a\right)^2}{\left(3c+a\right)\left(c+2d+a\right)}\)
\(\frac{d-a}{d+2a+b}+\frac{1}{2}=\frac{1}{2}.\frac{\left(3d+b\right)^2}{\left(3d+b\right)\left(d+2a+b\right)}\)
Cộng vế theo vế ta được
\(\frac{a-b}{a+2b+c}+\frac{1}{2}+\frac{b-c}{b+2c+d}+\frac{1}{2}+\frac{c-d}{c+2d+a}+\frac{1}{2}+\frac{d-a}{d+2a+b}+\frac{1}{2}=\frac{1}{2}.\frac{\left(3d+b\right)^2}{\left(3d+b\right)\left(d+2a+b\right)}+\frac{1}{2}.\frac{\left(3c+a\right)^2}{\left(3c+a\right)\left(c+2d+a\right)}+\frac{1}{2}.\frac{\left(3b+d\right)^2}{\left(3b+d\right)\left(b+2c+d\right)}+\frac{1}{2}.\frac{\left(3a+c\right)^2}{\left(3a+c\right)\left(a+2b+c\right)}\)
\(\ge\frac{1}{2}.\frac{\left(3a+c+3b+d+3c+a+3d+b\right)^2}{\left(3a+c\right)\left(a+2b+c\right)+\left(3b+d\right)\left(b+2c+d\right)+\left(3c+a\right)\left(c+2d+a\right)+\left(3d+b\right)\left(d+2a+b\right)}\)
\(=\frac{1}{2}.\frac{16\left(a+b+c+d\right)^2}{4\left(a+b+c+d\right)^2}=2\)
\(\Rightarrow A+2\ge2\)
\(\Leftrightarrow A\ge0\)
=4(a+b+c+d)2
Ta có: a−ba+2b+c +12 =12 .3a+ca+2b+c =12 .(3a+c)2(3a+c)(a+2b+c)
Tương tự ta có
b−cb+2c+d +12 =12 .(3b+d)2(3b+d)(b+2c+d)
c−dc+2d+a +12 =12 .(3c+a)2(3c+a)(c+2d+a)
d−ad+2a+b +12 =12 .(3d+b)2(3d+b)(d+2a+b)
Cộng vế theo vế ta được
a−ba+2b+c +12 +b−cb+2c+d +12 +c−dc+2d+a +12 +d−ad+2a+b +12 =12 .(3d+b)2(3d+b)(d+2a+b) +12 .(3c+a)2(3c+a)(c+2d+a) +12 .(3b+d)2(3b+d)(b+2c+d) +12 .(3a+c)2(3a+c)(a+2b+c)
≥12 .(3a+c+3b+d+3c+a+3d+b)2(3a+c)(a+2b+c)+(3b+d)(b+2c+d)+(3c+a)(c+2d+a)+(3d+b)(d+2a+b)
=12 .16(a+b+c+d)24(a+b+c+d)2 =2
⇒A+2≥2
a)
Do a,b,c > 0
nên áp dụng BĐT Svacxo ta được :
\(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\ge\frac{\left(a+b+c\right)^2}{a+b+c}=a+b+c\) ( đpcm )
Dấu '=' xảy ra \(\Leftrightarrow a=b=c\)
b)
Do a,b,c > 0
nên áp dụng BĐT Svacxo ta được :
\(\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\ge\frac{\left(a+b+c\right)^2}{b+c+c+a+a+b}=\frac{\left(a+b+c\right)^2}{2\left(a+b+c\right)}=\frac{a+b+c}{2}\) ( đpcm )
Dấu '=' xảy ra \(\Leftrightarrow a=b=c\)
trong sách nâng cao toán 8 của vũ hữu bình ấy bạn
giả sử \(a\ge b\ge c>0\)
Ta có : \(\frac{a^2}{b^2+c^2}-\frac{a}{b+c}=\frac{a\left(ab+ac-b^2-c^2\right)}{\left(b^2+c^2\right)\left(b+c\right)}=\frac{ab\left(a-b\right)+ac\left(a-c\right)}{\left(b^2+c^2\right)\left(b+c\right)}\)
TT: \(\frac{b^2}{c^2+a^2}-\frac{b}{c+a}=\frac{bc\left(b-c\right)+ba\left(b-a\right)}{\left(c^2+a^2\right)\left(c+a\right)}\)
\(\frac{c^2}{a^2+b^2}-\frac{c}{a+b}=\frac{ca\left(c-a\right)+cb\left(c-b\right)}{\left(a^2+b^2\right)\left(a+b\right)}\)
Do đó: \(\left(\frac{a^2}{b^2+c^2}+\frac{b^2}{c^2+a^2}+\frac{c^2}{a^2+b^2}\right)-\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)\)
\(=ab\left(a-b\right)\left[\frac{1}{\left(b^2+c^2\right)\left(b+c\right)}-\frac{1}{\left(c^2+a^2\right)\left(c+a\right)}\right]\)
\(+ca\left(a-c\right)\left[\frac{1}{\left(b^2+c^2\right)\left(b+c\right)}-\frac{1}{\left(a^2+b^2\right)\left(a+b\right)}\right]\)
\(+bc\left(b-c\right)\left[\frac{1}{\left(c^2+a^2\right)\left(c+a\right)}-\frac{1}{\left(a^2+b^2\right)\left(a+b\right)}\right]\)
Vì \(a\ge b\ge c\) => gtri bt > 0
=> đpcm