cho a,b,c là 3 số dương thoã mãn:a+b+c=1.Chứng minh rằng;
\(\dfrac{c+ab}{a+b}\)+\(\dfrac{a+bc}{b+c}\)+\(\dfrac{b+ac}{a+c}\)≥2
giải thích cho mình kĩ phần rút gọn
cho a b c là 3 số dương thoã mãn a+b+c=1 chứng minh rằng:
\(\dfrac{c+ab}{a+b}\)+\(\dfrac{a+bc}{b+c}\)+\(\dfrac{b+ac}{a+c}\)≥2
Đặt vế trái là P
\(P=\dfrac{1.c+ab}{a+b}+\dfrac{1.a+bc}{b+c}+\dfrac{1.b+ac}{a+c}=\dfrac{c\left(a+b+c\right)+ab}{a+b}+\dfrac{a\left(a+b+c\right)+bc}{b+c}+\dfrac{b\left(a+b+c\right)+ac}{a+c}\)
\(P=\dfrac{ac+c^2+bc+ab}{a+b}+\dfrac{a^2+ac+ab+bc}{b+c}+\dfrac{ab+ac+b^2+bc}{a+c}\)
\(P=\dfrac{c\left(a+c\right)+b\left(a+c\right)}{a+b}+\dfrac{a\left(a+c\right)+b\left(a+c\right)}{b+c}+\dfrac{a\left(b+c\right)+b\left(b+c\right)}{a+c}\)
\(P=\dfrac{\left(a+c\right)\left(b+c\right)}{a+b}+\dfrac{\left(a+b\right)\left(a+c\right)}{b+c}+\dfrac{\left(a+b\right)\left(b+c\right)}{a+c}\)
Áp dụng BĐT Cô-si:
\(\dfrac{\left(a+c\right)\left(b+c\right)}{a+b}+\dfrac{\left(a+b\right)\left(a+c\right)}{b+c}\ge2\sqrt{\dfrac{\left(a+c\right)\left(b+c\right)\left(a+b\right)\left(a+c\right)}{\left(a+b\right)\left(b+c\right)}}=2\left(a+c\right)\) (1)
Tương tự: \(\dfrac{\left(a+c\right)\left(b+c\right)}{a+b}+\dfrac{\left(a+b\right)\left(b+c\right)}{a+c}\ge2\left(b+c\right)\) (2)
\(\dfrac{\left(a+b\right)\left(a+c\right)}{b+c}+\dfrac{\left(a+b\right)\left(b+c\right)}{a+c}\ge2\left(a+b\right)\) (3)
Cộng vế với vế (1);(2);(3):
\(2.\dfrac{\left(a+b\right)\left(a+c\right)}{b+c}+2.\dfrac{\left(a+b\right)\left(b+c\right)}{a+c}+2.\dfrac{\left(a+c\right)\left(b+c\right)}{a+b}\ge2\left(a+b\right)+2\left(b+c\right)+2\left(c+a\right)\)
\(\Leftrightarrow\dfrac{\left(a+b\right)\left(a+c\right)}{b+c}+\dfrac{\left(a+b\right)\left(b+c\right)}{a+c}+\dfrac{\left(a+c\right)\left(b+c\right)}{a+c}\ge2\left(a+b+c\right)=2\) (đpcm)
Dấu "=" xảy ra khi \(a=b=c=\dfrac{1}{3}\)
Cho a,b,c là số thực dương thỏa mãn:a+b+c=1
Chứng minh rằng:\(\dfrac{a+bc}{b+c}+\dfrac{b+ca}{c+a}+\dfrac{c+ab}{a+b}\ge2\)
\(VT=\dfrac{a\left(a+b+c\right)+bc}{b+c}+\dfrac{b\left(a+b+c\right)+ca}{c+a}+\dfrac{c\left(a+b+c\right)+ab}{a+b}\)
\(VT=\dfrac{\left(a+b\right)\left(a+c\right)}{b+c}+\dfrac{\left(a+b\right)\left(b+c\right)}{c+a}+\dfrac{\left(a+c\right)\left(b+c\right)}{a+b}\)
Ta có:
\(\dfrac{\left(a+b\right)\left(a+c\right)}{b+c}+\dfrac{\left(a+b\right)\left(b+c\right)}{c+a}\ge2\left(a+b\right)\)
Tương tự: \(\dfrac{\left(a+b\right)\left(a+c\right)}{b+c}+\dfrac{\left(a+c\right)\left(b+c\right)}{a+b}\ge2\left(a+c\right)\)
\(\dfrac{\left(a+b\right)\left(b+c\right)}{a+c}+\dfrac{\left(a+c\right)\left(b+c\right)}{a+b}\ge2\left(b+c\right)\)
Cộng vế với vế:
\(\Rightarrow VT\ge2\left(a+b+c\right)=2\)
Dấu "=" xảy ra khi \(a=b=c=\dfrac{1}{3}\)
Cho a, b, c là các số thực dương thỏa mãn: a+b+c+ab+bc+ac=6. Chứng minh rằng: \(\dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a}\ge3\)
Ta có:
\(\left(a^2+1\right)+\left(b^2+1\right)+\left(c^2+1\right)+\left(a^2+b^2\right)+\left(b^2+c^2\right)+\left(c^2+a^2\right)\)
\(\ge2a+2b+2c+2ab+2bc+2ca=12\)
\(\Rightarrow3\left(a^2+b^2+c^2\right)+3\ge12\)
\(\Rightarrow a^2+b^2+c^2\ge3\)
\(P=\dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a}=\dfrac{a^4}{ab}+\dfrac{b^4}{bc}+\dfrac{c^4}{ca}\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{ab+bc+ca}\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{a^2+b^2+c^2}\)
\(P\ge a^2+b^2+c^2\ge3\)
\(P_{min}=3\) khi \(a=b=c=1\)
Cho a,b,c là ba số thực dương thỏa mãn điều kiện ab+bc+ac=3abc. Chứng minh rằng:
\(\sqrt{\dfrac{ab}{a+b+1}}+\sqrt{\dfrac{bc}{b+c+1}}+\sqrt{\dfrac{ca}{c+a+1}}\ge\sqrt{3}\)
1.Cho 3 số dương a,b,c. Chứng minh rằng:
\(\dfrac{19b^3-a^3}{ab+5b^2}+\dfrac{19c^3-b^3}{bc+5c^2}+\dfrac{19a^3-c^3}{ac+5a^2}\)≤ 3(a+b+c)
2.cho a,b,c dương thỏa man: a2+b2+c2=1
Tìm giá trị nhỏ nhất của biểu thức: P=\(\dfrac{bc}{a}+\dfrac{ac}{b}+\dfrac{ab}{c}\)
Cho a,b,c là 3 số dương thỏa mãn a+b+c=3. Chứng minh rằng :\(\dfrac{\sqrt{3a+bc}}{a+\sqrt{3a+bc}}+\dfrac{\sqrt{3b+ac}}{b+\sqrt{3b+ac}}+\dfrac{\sqrt{3c+ab}}{c+\sqrt{3c+ab}}\)≥ 2
Cho a,b,c là ba số thực dương thỏa mãn điều kiện ab+bc+ac=3abc. Chứng minh rằng:
\(\sqrt{\dfrac{ab}{a+b+1}}+\sqrt{\dfrac{bc}{b+c+1}}+\sqrt{\dfrac{ca}{c+a+1}}\ge\sqrt{3}\)
#Toán lớp 9
Cho a,b,c là ba số dương thỏa mãn a + b +c = 3 . Chứng minh rằng : \(\dfrac{\sqrt{3a+bc}}{a+\sqrt{3a+bc}}+\dfrac{\sqrt{3b+ac}}{b+\sqrt{3b+ac}}+\dfrac{\sqrt{3c+ab}}{c+\sqrt{3c+ab}}\) ≥ 2
Cho 3 số dương a, b, c thoã mãn a+b+c=1. Chứng minh rằng:
\(\sqrt{\dfrac{ab}{c + ab}} + \sqrt{\dfrac{bc}{a + bc}} + \sqrt{\dfrac{ca}{b + ac}} ≤ \dfrac{3}{2}\)
Lời giải:
Do $a+b+c=1$ nên:
\(\text{VT}=\sqrt{\frac{ab}{c(a+b+c)+ab}}+\sqrt{\frac{bc}{a(a+b+c)+bc}}+\sqrt{\frac{ca}{b(a+b+c)+ac}}\)
\(=\sqrt{\frac{ab}{(c+a)(c+b)}}+\sqrt{\frac{bc}{(a+b)(a+c)}}+\sqrt{\frac{ca}{(b+c)(b+a)}}\)
Áp dụng BĐT AM-GM:
\(\sqrt{\frac{ab}{(c+a)(c+b)}}\leq \frac{1}{2}\left(\frac{a}{c+a}+\frac{b}{c+b}\right)\)
\(\sqrt{\frac{bc}{(a+b)(a+c)}}\leq \frac{1}{2}\left(\frac{b}{a+b}+\frac{c}{c+a}\right)\)
\(\sqrt{\frac{ca}{(b+c)(b+a)}}\leq \frac{1}{2}\left(\frac{c}{b+c}+\frac{a}{b+a}\right)\)
Cộng theo vế:
\(\Rightarrow \text{VT}\leq \frac{1}{2}\left(\frac{a+b}{a+b}+\frac{b+c}{b+c}+\frac{c+a}{c+a}\right)=\frac{3}{2}\) (đpcm)
Dấu "=" xảy ra khi $a=b=c=\frac{1}{3}$