HSG Phú Thọ 2016
Cho các số dương a,b,c thỏa mãn \(ab+bc+ca=1\). Chứng minh rằng :
\(\frac{a-b}{1+c^2}+\frac{b-c}{1+a^2}+\frac{c-a}{1+b^2}=0\)
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G
22 tháng 7 2019
\(\frac{a+bc}{b+c}+\frac{b+ac}{c+a}+\frac{c+ab}{a+b}\)
\(=\frac{a\left(a+b+c\right)+bc}{b+c}+\frac{b\left(a+b+c\right)+ac}{a+c}+\frac{c\left(a+b+c\right)+ab}{a+b}\)
\(=\frac{\left(a+b\right)\left(a+c\right)}{b+c}+\frac{\left(a+b\right)\left(b+c\right)}{a+c}+\frac{\left(c+a\right)\left(c+b\right)}{a+b}\)
Áp dụng bđt Cô Si: \(\frac{\left(a+b\right)\left(a+c\right)}{b+c}+\frac{\left(a+b\right)\left(b+c\right)}{a+c}\ge2\left(a+b\right)\)
Tương tự,cộng theo vế và rút gọn =>đpcm
2 tháng 4 2020
\(\frac{a+bc}{b+c}+\frac{b+ac}{c+a}+\frac{c+ab}{a+b}\)
\(=\frac{a\left(a+b+c\right)+bc}{b+c}+\frac{b\left(a+b+c\right)+ac}{a+c}+\frac{c\left(a+b+c\right)+ab}{a+b}\)
\(=\frac{\left(a+b\right)\left(a+c\right)}{b+c}+\frac{\left(a+b\right)\left(b+c\right)}{a+c}+\frac{\left(c+a\right)\left(c+b\right)}{a+b}\)
Áp dụng bđt CÔ si
\(\frac{\left(a+b\right)\left(a+c\right)}{b+c}+\frac{\left(a+b\right)\left(b+c\right)}{a+c}\ge2\left(a+b\right)\)
.............
8 tháng 8 2021
Bài 1:
Ta có: a + b - 2c = 0
⇒ a = 2c − b thay vào a2 + b2 + ab - 3c2 = 0 ta có:
(2c − b)2 + b2 + (2c − b).b − 3c2 = 0
⇔ 4c2 − 4bc + b2 + b2 + 2bc − b2 − 3c2 = 0
⇔ b2 − 2bc + c2 = 0
⇔ (b − c)2 = 0
⇔ b − c = 0
⇔ b = c
⇒ a + c − 2c = 0
⇔ a − c = 0
⇔ a = c
⇒ a = b = c
Vậy a = b = c
T
24 tháng 11 2019
Bài 1:
Ta có: \(\frac{ab}{a+b}=ab.\frac{1}{a+b}\le\frac{ab}{4}\left(\frac{1}{a}+\frac{1}{b}\right)=\frac{b}{4}+\frac{a}{4}\)
Tương tự các BĐT còn lại rồi cộng theo vế ta có d9pcm.
Bài 2: 2 bài đều dùng Svac cả!
(cần thì ib t gửi link cho)
21 tháng 10 2018
Ta có:
\(M=\frac{19a+3}{1+b^2}+\frac{19b+3}{c^2+1}+\frac{19c+3}{a^2+1}\)
\(=19a-\frac{19ab^2-3}{b^2+1}+19b-\frac{19bc^2-3}{c^2+1}+\frac{19ca^2-3}{a^2+1}\)
\(\ge19\left(a+b+c\right)-\frac{19ab^2-3}{2b}-\frac{19bc^2-3}{2c}-\frac{19ca^2-3}{2a}\)
\(=19\left(a+b+c\right)-19\left(\frac{ab}{2}+\frac{bc}{2}+\frac{ca}{2}\right)+\frac{3}{2}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
\(\ge19.3-\frac{19.3}{2}+\frac{3}{2}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=\frac{19.3}{2}+\frac{3}{2}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
Lại có:
\(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2\ge3\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)\ge3\frac{\left(1+1+1\right)^2}{ab+bc+ca}=\frac{3.9}{3}=9\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge3\)
\(\Rightarrow M\ge\frac{19.3}{2}+\frac{3}{2}.3=33\)
\(\)
26 tháng 7 2021
Vì \(\left|a\right|\le1;\left|b-1\right|\le2\)
\(=>\left|a\right|\cdot\left|b-1\right|=\left|ab-a\right|\le2\)
Áp dụng BĐT \(\left|x\right|+\left|y\right|\ge\left|x+y\right|\) ta có:
\(\left|a-c+ab-a\right|\le\left|a-c\right|+\left|ab-a\right|=2+3=5\)
\(=>\left|ab-c\right|\le5\)
NV
Nguyễn Việt Lâm
Giáo viên
13 tháng 5 2019
\(P=\left(b+c+d\right)\left(\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\right)=1+\frac{b}{c}+\frac{b}{d}+\frac{c}{b}+1+\frac{c}{d}+\frac{d}{b}+\frac{d}{c}+1\)
\(=3+\frac{b}{c}+\frac{b}{d}+\frac{c}{d}+\frac{c}{b}+\frac{d}{b}+\frac{d}{c}\)
Mặt khác do \(b\le c\le d\Rightarrow\left(d-c\right)\left(c-b\right)\ge0\)
\(\Leftrightarrow cd-bd-c^2+bc\ge0\Leftrightarrow bc+cd\ge c^2+bd\)
\(\Leftrightarrow\frac{bc+cd}{cd}\ge\frac{c^2+bd}{cd}\Leftrightarrow\frac{b}{d}+1\ge\frac{c}{d}+\frac{b}{c}\)
\(\frac{bc+cd}{bc}\ge\frac{c^2+bd}{bc}\Leftrightarrow\frac{d}{b}+1\ge\frac{c}{b}+\frac{d}{c}\)
\(\Leftrightarrow\frac{b}{d}+\frac{d}{b}+2\ge\frac{b}{c}+\frac{c}{d}+\frac{c}{b}+\frac{d}{c}\)
\(\Leftrightarrow2\left(\frac{b}{d}+\frac{d}{b}\right)+2\ge\frac{b}{c}+\frac{b}{d}+\frac{c}{d}+\frac{c}{b}+\frac{d}{b}+\frac{d}{c}=P\)
Mà \(a\le b\le d\le2a\Rightarrow\left\{{}\begin{matrix}\frac{1}{2}\le\frac{b}{d}\le1\\1\le\frac{d}{b}\le2\end{matrix}\right.\)
\(\Rightarrow\left(\frac{b}{d}-1\right)\left(\frac{d}{b}-2\right)\ge0\Leftrightarrow1-2\frac{b}{d}-\frac{d}{b}+2\ge0\)
\(\Leftrightarrow\frac{b}{d}+\frac{d}{b}\le3-\frac{b}{d}\le3-\frac{1}{2}=\frac{5}{2}\)
\(\Rightarrow P\le2.\frac{5}{2}+2=7\)
Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}b=c=a\\d=2a\end{matrix}\right.\)
16 tháng 10 2018
\(\frac{a^2}{b}-a+b+b=\frac{a^2-ab+b^2}{b}+b\ge2\sqrt{a^2-ab+b^2}\)
\(=\sqrt{a^2-ab+b^2}+\sqrt{a^2-ab+b^2}=\sqrt{a^2-ab+b^2}+\sqrt{\frac{3}{4}\left(a-b\right)^2+\frac{1}{4}\left(a+b\right)^2}\)
\(\ge\sqrt{a^2-ab+b^2}+\sqrt{\frac{1}{4}\left(a+b\right)^2}=\sqrt{a^2-ab+b^2}+\frac{a+b}{2}\)
chứng minh tương tự ta được
\(\frac{b^2}{c}-b+c+c\ge\sqrt{b^2-bc+c^2}+\frac{b+c}{2},\frac{c^2}{a}-c+a+a\ge\sqrt{c^2-ca+a^2}+\frac{a+c}{2}\)
cộng vế với vế ta được
\(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}+a+b+c\ge\sqrt{a^2-ab+b^2}+\sqrt{b^2-bc+c^2}+\sqrt{c^2-ca+a^2}+a+b+c\)
Dấu bằng xảy ra khi a=b=c
Ta có 1+c2=ab+bc+ca+c2=(a+c)(b+c)
Tương tự 1+a2=(a+b)(a+c)
1+b2=(a+b)(b+c)
Suy ra \(\frac{a-b}{1+c^2}=\frac{a-b}{\left(a+c\right)\left(b+c\right)}=\frac{1}{c+b}-\frac{1}{c+a}\)
\(\frac{b-c}{1+a^2}=\frac{b-c}{\left(a+b\right)\left(a+c\right)}=\frac{1}{a+c}-\frac{1}{a+b}\)
\(\frac{c-a}{1+b^2}=\frac{c-a}{\left(a+b\right)\left(b+c\right)}=\frac{1}{a+b}-\frac{1}{b+c}\)
\(\Rightarrow\frac{a-b}{1+c^2}+\frac{b-c}{1+a^2}+\frac{c-a}{1+b^2}=\frac{1}{c+b}-\frac{1}{c+a}+\frac{1}{a+c}-\frac{1}{a+b}+\frac{1}{a+b}-\frac{1}{b+c}=0\)