Chứng minh rằng với mọi số thực a, b, c dương thỏa mãn a2 + b2 + c2 = 27. CMR:
\(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=\frac{12}{a^2+63}+\frac{12}{b^2+63}+\frac{12}{c^2+63}\)
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4 tháng 12 2019
Cần CM: \(\frac{1}{9-a}-\frac{12}{a^2+63}\ge\frac{1}{144}a^2-\frac{1}{16}\) (1)
\(\Leftrightarrow\)\(\frac{a^2+12a-45}{\left(9-a\right)\left(a^2+63\right)}\ge\frac{1}{144}a^2-\frac{1}{16}\)
\(\Leftrightarrow\)\(144\left(a^2+12a-45\right)\ge\left(a-3\right)\left(a+3\right)\left(9-a\right)\left(a^2+63\right)\)
\(\Leftrightarrow\)\(\left(a-3\right)\left[144\left(a+15\right)-\left(a+3\right)\left(9-a\right)\left(a^2+63\right)\right]\ge0\)
\(\Leftrightarrow\)\(\left(a-3\right)\left(a^4-6a^3+36a^2-234a+459\right)\ge0\)
\(\Leftrightarrow\)\(\left(a-3\right)^2\left(a^3-3a^2+27a+153\right)\ge0\)
\(\Leftrightarrow\)\(\left(a-3\right)^2\left[\left(a-3\right)^2\left(a+3\right)+36a+126\right]\ge0\) ( đúng )
Do đó (1) đúng => \(\Sigma_{cyc}\frac{1}{9-a}-\Sigma_{cyc}\frac{12}{a^2+63}\ge\frac{1}{144}\left(a^2+b^2+c^2\right)-\frac{3}{16}=0\)
\(\Rightarrow\)\(\Sigma_{cyc}\frac{12}{a^2+63}\le\Sigma_{cyc}\frac{1}{9-a}\le\Sigma_{cyc}\frac{1}{a+b}\) ( do \(a+b+c\le9\) )
Dấu "=" xảy ra khi a=b=c=3
9 tháng 8 2021
Ta có: \(2a+b^2=2a\left(a+b+c\right)+b^2=b^2+2a^2+2ab+2ac\)
\(\ge4ab+2ac+a^2\)
\(\Rightarrow\frac{a}{2a+b^2}\le\frac{a}{4ab+2ac+a^2}=\frac{1}{4b+2c+a}\)
\(\le\frac{1}{49}.\frac{49}{4b+2c+a}=\frac{1}{49}.\frac{\left(4+2+1\right)^2}{4b+2c+a}\)
\(\le\frac{1}{49}\left(\frac{16}{4b}+\frac{4}{2c}+\frac{1}{a}\right)=\frac{1}{49}\left(\frac{4}{b}+\frac{2}{c}+\frac{1}{a}\right)\)
CMTT: \(\frac{b}{2b+c^2}\le\frac{1}{49}\left(\frac{4}{c}+\frac{2}{a}+\frac{1}{b}\right);\frac{c}{2c+a^2}\le\frac{1}{49}\left(\frac{4}{a}+\frac{2}{b}+\frac{1}{c}\right)\)
\(\Rightarrow\frac{a}{2a+b^2}+\frac{b}{2b+c^2}+\frac{c}{2c+a^2}\le\frac{1}{7}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)( đpcm )
NV
Nguyễn Việt Lâm
Giáo viên
30 tháng 12 2020
1. Đề thiếu
2. BĐT cần chứng minh tương đương:
\(a^4+b^4+c^4\ge abc\left(a+b+c\right)\)
Ta có:
\(a^4+b^4+c^4\ge\dfrac{1}{3}\left(a^2+b^2+c^2\right)^2\ge\dfrac{1}{3}\left(ab+bc+ca\right)^2\ge\dfrac{1}{3}.3abc\left(a+b+c\right)\) (đpcm)
3.
Ta có:
\(\left(a^6+b^6+1\right)\left(1+1+1\right)\ge\left(a^3+b^3+1\right)^2\)
\(\Rightarrow VT\ge\dfrac{1}{\sqrt{3}}\left(a^3+b^3+1+b^3+c^3+1+c^3+a^3+1\right)\)
\(VT\ge\sqrt{3}+\dfrac{2}{\sqrt{3}}\left(a^3+b^3+c^3\right)\)
Lại có:
\(a^3+b^3+1\ge3ab\) ; \(b^3+c^3+1\ge3bc\) ; \(c^3+a^3+1\ge3ca\)
\(\Rightarrow2\left(a^3+b^3+c^3\right)+3\ge3\left(ab+bc+ca\right)=9\)
\(\Rightarrow a^3+b^3+c^3\ge3\)
\(\Rightarrow VT\ge\sqrt{3}+\dfrac{6}{\sqrt{3}}=3\sqrt{3}\)
NV
Nguyễn Việt Lâm
Giáo viên
30 tháng 12 2020
4.
Ta có:
\(a^3+1+1\ge3a\) ; \(b^3+1+1\ge3b\) ; \(c^3+1+1\ge3c\)
\(\Rightarrow a^3+b^3+c^3+6\ge3\left(a+b+c\right)=9\)
\(\Rightarrow a^3+b^3+c^3\ge3\)
5.
Ta có:
\(\dfrac{a}{b}+\dfrac{b}{c}\ge2\sqrt{\dfrac{a}{c}}\) ; \(\dfrac{a}{b}+\dfrac{c}{a}\ge2\sqrt{\dfrac{c}{b}}\) ; \(\dfrac{b}{c}+\dfrac{c}{a}\ge2\sqrt{\dfrac{b}{a}}\)
\(\Rightarrow\sqrt{\dfrac{b}{a}}+\sqrt{\dfrac{c}{b}}+\sqrt{\dfrac{a}{c}}\le\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}=1\)
AH
Akai Haruma
Giáo viên
5 tháng 4 2021
Lời giải:
Áp dụng BĐT AM-GM:
\(\text{VT}=\sum \frac{a+1}{b^2+1}=\sum [(a+1)-\frac{b^2(a+1)}{b^2+1}]=\sum (a+1)-\sum \frac{b^2(a+1)}{b^2+1}\)
\(=6-\sum \frac{b^2(a+1)}{b^2+1}\geq 6-\sum \frac{b^2(a+1)}{2b}=6-\sum \frac{ab+b}{2}\)
\(=6-\frac{\sum ab+3}{2}\geq 6-\frac{\frac{1}{3}(a+b+c)^2+3}{2}=6-\frac{3+3}{2}=3\)
Ta có đpcm
Dấu "=" xảy ra khi $a=b=c=1$
30 tháng 5 2018
Ta có; \(\frac{a^2}{a+b}+\frac{a+b}{4}\ge2\sqrt{\frac{a^2}{a+b}.\frac{a+b}{4}}=a\)
Tương tự : \(\frac{b^2}{b+c}+\frac{b+c}{4}\ge b\)
\(\frac{c^2}{c+a}+\frac{c+a}{4}\ge c\)
Cộng từng vế ta có:
\(\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+a}+\frac{a+b+c}{2}\ge a+b+c\)
\(\Leftrightarrow\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+a}\ge\frac{a+b+c}{2}=\frac{1}{2}\)
19 tháng 12 2018
Áp dụng BĐT AM-GM: \(1+b^2\ge2b\)
\(\Rightarrow\frac{a}{1+b^2}=a-\frac{ab^2}{1+b^2}\ge a-\frac{ab^2}{2b}=a-\frac{ab}{2}\)
Tương tự: \(\frac{b}{1+c^2}\ge b-\frac{bc}{2};\frac{c}{1+a^2}\ge c-\frac{ca}{2}\)
Cộng vế với vế 3 BĐT trên ta được: \(\frac{a}{1+b^2}+\frac{b}{1+c^2}+\frac{c}{1+a^2}\ge\left(a+b+c\right)-\frac{ab+bc+ca}{2}=3-\frac{ab+bc+ca}{2}\)
Mà \(ab+bc+ca\le\frac{\left(a+b+c\right)^2}{3}\)
Nên \(\frac{a}{1+b^2}+\frac{b}{1+c^2}+\frac{c}{1+a^2}\ge3-\frac{\left(a+b+c\right)^2}{6}=3-\frac{9}{6}=\frac{3}{2}\)(đpcm).
Dấu "=" xảy ra <=> a=b=c=1.
Ta có: \(\frac{1}{a+b}+\frac{1}{b+c}\ge2\sqrt{\frac{1}{a+b}\frac{1}{b+c}}=2\frac{1}{\sqrt{\left(a+b\right)\left(b+c\right)}}\ge\frac{4}{a+2b+c}\)
Tương tự có: \(\frac{1}{b+c}+\frac{1}{a+c}\ge\frac{4}{a+2c+b}\)
\(\frac{1}{a+b}+\frac{1}{a+c}\ge\frac{4}{b+2a+c}\)
\(\Rightarrow\frac{1}{a+b}+\frac{1}{c+b}+\frac{1}{a+c}\ge2\left(\frac{1}{b+2a+c}+\frac{1}{a+2b+c}+\frac{1}{b+2c+a}\right)\)
Ta CM: \(\frac{1}{b+2a+c}\ge\frac{6}{a^2+63}\). Thật vậy:
\(\frac{1}{b+2a+c}\ge\frac{6}{a^2+63}\)\(\Leftrightarrow a^2+63\ge6b+12a+6c\)\(\Leftrightarrow2a^2+b^2+c^2+36-6b-12a-6c\ge0\)
\(\Leftrightarrow2\left(a-3\right)^2+\left(b-3\right)^2+\left(c-3\right)^2\ge0\) ( luôn đúng)
Dấu '=' xảy ra <=> a=b=c=3
Vậy \(\frac{1}{b+2a+c}+\frac{1}{a+2b+c}+\frac{1}{b+2c+a}\ge\frac{6}{a^2+63}+\frac{6}{b^2+63}+\frac{6}{c^2+63}\)
=> đpcm