\(a,b,c>0\) , \(a+b+c=3\), chứng minh \(P=a^3+b^3+c^3+ab+bc+ca\ge6\)
K
Khách
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M
0
MK
1
7 tháng 7 2019
\(\frac{a-bc}{a+bc}=\frac{a-bc}{a\left(a+b+c\right)+bc}=\frac{a-bc}{a^2+ab+bc+ca}=\frac{a-bc}{\left(a+b\right)\left(c+a\right)}\)
\(=\left(a-bc\right)\sqrt{\frac{1}{\left(a+b\right)^2\left(c+a\right)^2}}\le\frac{\frac{a-bc}{\left(a+b\right)^2}+\frac{a-bc}{\left(c+a\right)^2}}{2}=\frac{a-bc}{2\left(a+b\right)^2}+\frac{a-bc}{2\left(c+a\right)^2}\)
Tương tự, ta có: \(\frac{b-ca}{b+ca}\le\frac{b-ca}{2\left(b+c\right)^2}+\frac{b-ca}{2\left(a+b\right)^2}\)\(;\)\(\frac{c-ab}{c+ab}\le\frac{c-ab}{2\left(c+a\right)^2}+\frac{c-ab}{2\left(b+c\right)^2}\)
=> \(\frac{a-bc}{a+bc}+\frac{b-ca}{b+ca}+\frac{c-ab}{c+ab}\le\frac{a-bc+b-ca}{2\left(a+b\right)^2}+\frac{b-ca+c-ab}{2\left(b+c\right)^2}+\frac{a-bc+c-ab}{2\left(c+a\right)^2}\)
\(\frac{\left(a+b\right)\left(1-c\right)}{2\left(a+b\right)\left(1-c\right)}+\frac{\left(b+c\right)\left(1-a\right)}{2\left(b+c\right)\left(1-a\right)}+\frac{\left(c+a\right)\left(1-b\right)}{2\left(c+a\right)\left(1-b\right)}=\frac{3}{2}\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(a=b=c=\frac{1}{3}\)
NV
Nguyễn Việt Lâm
Giáo viên
25 tháng 3 2019
Biến đổi tương đương:
\(\left(a+b+c\right)^2\ge3\left(ab+ac+bc\right)\)
\(\Leftrightarrow a^2+b^2+c^2+2ab+2ac+2bc\ge3\left(ab+ac+bc\right)\)
\(\Leftrightarrow a^2+b^2+c^2-ab-ac-bc\ge0\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2ac-2bc\ge0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(a-c\right)^2+\left(b-c\right)^2\ge0\) (luôn đúng)
Dấu "=" xảy ra khi \(a=b=c\)
\(\Rightarrow\frac{\left(a+b+c\right)^2}{ab+ac+bc}\ge3\)
b/ \(VT=\frac{\left(a+b+c\right)^2}{ab+ac+bc}+\frac{ab+ac+bc}{\left(a+b+c\right)^2}=\frac{8\left(a+b+c\right)^2}{9\left(ab+ac+bc\right)}+\frac{\left(a+b+c\right)^2}{9\left(ab+ac+bc\right)}+\frac{ab+ac+bc}{\left(a+b+c\right)^2}\)
\(\Rightarrow VT\ge\frac{8\left(a+b+c\right)^2}{9\left(ab+ac+bc\right)}+2\sqrt{\frac{\left(a+b+c\right)^2\left(ab+ac+bc\right)}{9\left(ab+ac+bc\right)\left(a+b+c\right)^2}}\ge\frac{8.3}{9}+\frac{2}{3}=\frac{10}{3}\) (đpcm)
Dấu "=" xảy ra khi \(a=b=c\)
D
1
24 tháng 4 2019
Áp dụng bđt AM-GM:
\(\frac{a^3}{b}+ab\ge2a^2\)
\(\frac{b^3}{c}+bc\ge2b^2\)
\(\frac{c^3}{a}+ac\ge2c^2\)
\(\Rightarrow\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}+ab+bc+ac\ge2a^2+2b^2+2c^2\ge2ab+2ac+2bc\)
\(\Rightarrow\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}\ge ab+bc+ac\left(đpcm\right)\)
\("="\Leftrightarrow a=b=c\)
NT
3
19 tháng 3 2019
\(\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}=\frac{a^4}{ab}+\frac{b^4}{bc}+\frac{c^4}{ac}\ge\frac{\left(a^2+b^2+c^2\right)^2}{ab+bc+ac}=\frac{\left(ab+bc+ac\right)^2}{ab+bc+ca}=ab+bc+ac\)
\("="\Leftrightarrow a=b=c\)
NV
Nguyễn Việt Lâm
Giáo viên
19 tháng 3 2019
\(\frac{a^4}{ab}+\frac{b^4}{bc}+\frac{c^4}{ac}\ge\frac{\left(a^2+b^2+c^2\right)^2}{ab+ac+bc}\ge\frac{\left(ab+ac+bc\right)^2}{ab+ac+bc}=ab+ac+bc\) (đpcm)
Dấu "=" xảy ra khi \(a=b=c\)
Ta có:
\(ab+bc+ca\le\frac{\left(a+b+c\right)^2}{3}=3\)
\(\Rightarrow VT-VP=a^3+b^3+c^3+ab+bc+ca-6\ge a^3+b^3+c^3-ab-bc-ca\) (Giải thích:\(-6\ge-2\left(ab+bc+ca\right)\Rightarrow a^3+b^3+c^3+ab+bc+ca-6\ge a^3+b^3+c^3-ab-bc-ca\))
Ta lại có:
\(a^3+b^3+c^3-ab-bc-ca\ge\frac{\left(a^2+b^2+c^2\right)^2}{a+b+c}-\frac{\left(a+b+c\right)^2}{3}\ge\frac{\left[\frac{\left(a+b+c\right)^2}{3}\right]^2}{3}-3=0\)
\(\Rightarrow VT-VP\ge0\)
\(\Rightarrow P\ge6\)
Nếu có không đúng thì nhớ nói nhe chớ đừng có k sai tui giống mấy lần trước nhe :(
Bài ở dưới mình nhầm nhe.
Update
Ta có:
\(a^3+b^3+c^3\ge\frac{\left(a^2+b^2+c^2\right)^2}{a+b+c}=\frac{\left(a^2+b^2+c^2\right)\left(a^2+b^2+c^2\right)}{3}\ge\frac{\left(a^2+b^2+c^2\right)\frac{\left(a+b+c\right)^2}{3}}{3}=a^2+b^2+c^2\)
\(\Rightarrow P\ge a^2+b^2+c^2+ab+bc+ca=\frac{a^2+b^2+c^2}{2}+\frac{\left(a+b+c\right)^2}{2}\ge\frac{\frac{\left(a+b+c\right)^2}{3}}{2}+\frac{9}{2}=6\)