giải pt với b=2 :\(\frac{ab}{a+b}+\frac{bc}{b+c}+\frac{ca}{c+a}=\frac{a+b+c}{2}\).tìm a,c
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KN
15 tháng 1 2020
\(ĐKXĐ:a,b,c\ne0\)
\(\frac{x-a}{bc}+\frac{x-b}{ca}+\frac{x-c}{ab}=\frac{2}{a}+\frac{2}{b}+\frac{2}{c}\)
\(\Leftrightarrow\frac{xa-a^2}{abc}+\frac{xb-b^2}{abc}+\frac{xc-c^2}{abc}=\frac{2bc}{abc}+\frac{2ac}{abc}+\frac{2ab}{abc}\)
\(\Leftrightarrow\frac{xa-a^2+xb-b^2+xc-c^2}{abc}=\frac{2bc+2ac+2ab}{abc}\)
\(\Leftrightarrow xa-a^2+xb-b^2+xc-c^2=2bc+2ac+2ab\)
\(\Leftrightarrow xa+xb+xc=2bc+2ac+2ab+a^2+b^2+c^2\)
\(\Leftrightarrow x\left(a+b+c\right)=\left(a+b+c\right)^2\)
\(\Leftrightarrow x=a+b+c\)
Vậy x = a + b + c
KN
15 tháng 1 2020
\(ĐKXĐ:a,b,c\ne0\)
\(\frac{a+b-x}{c}+\frac{b+c-x}{a}+\frac{c+a-x}{b}+\frac{4x}{a+b+c}=1\)
\(\Leftrightarrow\frac{a+b-x}{c}+\frac{b+c-x}{a}+\frac{c+a-x}{b}=1-\frac{4x}{a+b+c}\)
\(\Leftrightarrow1+\frac{a+b-x}{c}+1+\frac{b+c-x}{a}+1+\frac{c+a-x}{b}=4\)
\(-\frac{4x}{a+b+c}\)
\(\Leftrightarrow\frac{a+b+c-x}{c}+\frac{a+b+c-x}{a}+\frac{a+b+c-x}{b}=\)
\(\frac{4\left(a+b+c\right)}{a+b+c}-\frac{4x}{a+b+c}\)
\(\Leftrightarrow\frac{a+b+c-x}{c}+\frac{a+b+c-x}{a}+\frac{a+b+c-x}{b}=\)
\(\frac{4\left(a+b+c-x\right)}{a+b+c}\)
\(\Leftrightarrow\left(a+b+c-x\right)\left(\frac{1}{c}+\frac{1}{a}+\frac{1}{b}-\frac{4}{a+b+c}\right)=0\)
\(\Rightarrow\left(a+b+c-x\right)=0\)hoặc \(\left(\frac{1}{c}+\frac{1}{a}+\frac{1}{b}-\frac{4}{a+b+c}\right)=0\)
+) Nếu \(\Rightarrow\left(a+b+c-x\right)=0\)thì x = a + b + c
+) Nếu \(\left(\frac{1}{c}+\frac{1}{a}+\frac{1}{b}-\frac{4}{a+b+c}\right)=0\)thì x thỏa mãn với mọi số
TH
23 tháng 3 2019
\(\frac{x-a}{bc}+\frac{x-b}{ac}+\frac{x-c}{ab}=2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
\(\frac{x-a}{bc}+\frac{x-b}{ac}+\frac{x-c}{ab}=\frac{2}{a}+\frac{2}{b}+\frac{2}{c}\)
\(\frac{ax-a^2+bx-b^2+cx-c^2}{abc}=2\left(\frac{ab+bc+ac}{abc}\right)\)
\(ax-a^2+bx-b^2+cx-c^2=2\left(ab+bc+ac\right)\)
\(x\left(a+b+c\right)-\left(a^2+b^2+c^2\right)=2\left(ab+bc+ac\right)\)
\(x\left(a+b+c\right)=a^2+b^2+c^2+2ab+2bc+2ac\)
\(x=a+b+c\)
19 tháng 11 2019
a)\(VT=\sum_{cyc}\frac{ab^3+ab^2c+a^2bc}{\left(a^2+bc+ca\right)\left(b^2+bc+ca\right)}\le\frac{\sum_{cyc}\left(ab^3+ab^2c+a^2bc\right)}{\left(ab+bc+ca\right)^2}\)
\(=\frac{ab^3+bc^3+ca^3+2a^2bc+2ab^2c+2abc^2}{\left(ab+bc+ca\right)^2}\)\(\le\frac{\sum_{cyc}ab\left(a^2+b^2\right)+abc\left(a+b+c\right)}{\left(ab+bc+ca\right)^2}\)
\(=\frac{\left(ab+bc+ca\right)\left(a^2+b^2+c^2\right)}{\left(ab+bc+ca\right)^2}=\frac{a^2+b^2+c^2}{ab+bc+ca}=VP\)
P
5 tháng 1 2018
\(\frac{bc+a^2}{a+b}+\frac{ac+b^2}{b+c}+\frac{ab+c^2}{a+c}\ge\)a+b+c
<=>\(\frac{bc+a^2}{a+b}-a+\frac{ac+b^2}{b+c}-b+\frac{ab+c^2}{a+c}-c\ge0\)
<=>\(\frac{b\left(c-a\right)}{a+b}+\frac{c\left(a-b\right)}{b+c}+\frac{a\left(b-c\right)}{a+c}\ge0\)
<=>\(\frac{b\left(b+c\right)\left(a+c\right)\left(a-c\right)}{\left(a+b\right)\left(c+c\right)\left(a+c\right)}\)+\(\frac{c\left(a+c\right)\left(a-b\right)\left(a+b\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)+\(\frac{a\left(a+b\right)\left(b-c\right)\left(b+c\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\ge0\)
<=>\(\frac{b^2c^2-b^2a^2+bc^3-a^2bc}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)+\(\frac{a^3c-ab^2c+c^2a^2-b^2c^2}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)+\(\frac{a^2b^2-a^2c^2+ab^3-abc^2}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\ge0\)
<=>\(\frac{bc^3+a^3c+ab^3-a^2bc-ab^2c-abc^2}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\ge0\)
<=>\(\frac{2bc^3+2a^3c+2ab^3-2a^2bc-2ab^2c-2abc^2}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)>=0
<=>\(\frac{bc\left(c-a\right)^2+ac\left(a-b\right)^2+ab\left(b-c\right)^2}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\ge0\)(đung voi moi a,b,c >0)
Dấu ''='' xay ra khi a=b=c
NV
Nguyễn Việt Lâm
Giáo viên
14 tháng 4 2022
Bunhiacopxki:
\(\left(a^2+bc+ca\right)\left(b^2+bc+ca\right)\ge\left(ab+bc+ca\right)^2\)
\(\Rightarrow\dfrac{ab}{a^2+bc+ca}\le\dfrac{ab\left(b^2+bc+ca\right)}{\left(ab+bc+ca\right)^2}\)
Tương tự: \(\dfrac{bc}{b^2+ca+ab}\le\dfrac{bc\left(c^2+ca+ab\right)}{\left(ab+bc+ca\right)^2}\)
\(\dfrac{ca}{c^2+ab+bc}\le\dfrac{ca\left(a^2+ab+bc\right)}{\left(ab+bc+ca\right)^2}\)
\(\Rightarrow VT\le\dfrac{ab\left(b^2+bc+ca\right)+bc\left(c^2+ca+ab\right)+ca\left(a^2+ab+bc\right)}{\left(ab+bc+ca\right)^2}\)
Nên ta chỉ cần chứng minh:
\(\dfrac{ab\left(b^2+bc+ca\right)+bc\left(c^2+ca+ab\right)+ca\left(a^2+ab+bc\right)}{\left(ab+bc+ca\right)^2}\le\dfrac{a^2+c^2+c^2}{ab+bc+ca}\)
\(\Leftrightarrow ab\left(b^2+bc+ca\right)+bc\left(c^2+ca+ab\right)+ca\left(a^2+ab+bc\right)\le\left(a^2+b^2+c^2\right)\left(ab+bc+ca\right)\)
Nhân phá và rút gọn 2 vế:
\(\Leftrightarrow a^3b+b^3c+c^3a\ge abc\left(a+b+c\right)\)
\(\Leftrightarrow\dfrac{a^3b+b^3c+c^3a}{abc}\ge a+b+c\)
\(\Leftrightarrow\dfrac{a^2}{c}+\dfrac{b^2}{a}+\dfrac{c^2}{b}\ge a+b+c\)
Đúng do: \(\dfrac{a^2}{c}+\dfrac{b^2}{a}+\dfrac{c^2}{b}\ge\dfrac{\left(a+b+c\right)^2}{a+b+c}=a+b+c\)
Dấu "=" xảy ra khi \(a=b=c\)
8 tháng 7 2019
\(\frac{a^2}{a+bc}=\frac{a^3}{a^2+abc}=\frac{a^3}{a^2+ab+bc+ac}=\frac{a^3}{\left(a+b\right)\left(a+c\right)}\)
Áp dụng BĐT cosi
\(\frac{a^3}{\left(a+b\right)\left(a+c\right)}+\frac{a+b}{8}+\frac{a+c}{8}\ge\frac{3}{4}a\)
Tương tự
=> \(A\ge\frac{3}{4}\left(a+b+c\right)-\frac{1}{2}\left(a+b+c\right)=\frac{1}{4}\left(a+b+c\right)\)
Lại có \(\left(a+b+c\right)\ge\frac{9}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}=\frac{9}{1}=9\)
=> \(A\ge\frac{9}{4}\)
MinA=9/4 khi a=b=c=3
23 tháng 8 2020
Bài 1: Ta có \(\left(\frac{a^2}{b}-a+b\right)+b^2=\frac{a^2-ab+b^2}{b}+b\ge2\sqrt{a^2-ab+b^2}\) (áp dụng Bất Đẳng Thức Cosi)
\(=\sqrt{a^2-ab+b^2}+\sqrt{\frac{3}{4}\left(a-b\right)^2+\frac{1}{4}\left(a+b\right)^2}\ge\sqrt{a^2-ab+b^2}+\frac{1}{2}\left(a+b\right)\)
\(\Rightarrow\frac{a^2}{b}-a+2b\ge\sqrt{a^2-ab+b^2}+\frac{1}{2}\left(a+b\right)\left(1\right)\)
Tương tự ta có \(\hept{\begin{cases}\frac{b^2}{c}-b+2c\ge\sqrt{b^2-bc+c^2}+\frac{1}{2}\left(b+c\right)\left(2\right)\\\frac{c^2}{a}-c+2a\ge\sqrt{c^2-ac+a^2}+\frac{1}{2}\left(a+c\right)\left(3\right)\end{cases}}\)
Từ (1) và (2) và (3) \(\Rightarrow\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\ge\sqrt{a^2-ab+b^2}+\sqrt{b^2-bc+c^2}+\sqrt{c^2-ac+a^2}\)
Dấu "=" xảy ra khi a=b=c
Ta có: \(\frac{ab}{a+b}\le\frac{a+b}{4};\frac{bc}{b+c}\le\frac{b+c}{4};\frac{ca}{c+a}\le\frac{c+a}{4}\)
\(\Rightarrow VT\le\frac{a+b+c}{2}\). Dấu = xảy ra khi a = b= c mà b =2 => a = c =2
Sử dụng phương pháp U.C.T