Cho 3 số dương a,b,c biết 0≤ a ≤ b ≤ c ≤ 1
Chứng minh rằng \(\dfrac{a}{bc+1}+\dfrac{b}{ac+1}+\dfrac{c}{ab+1}\) ≤ 2
Cho 3 số dương a,b,c biết 0≤ a ≤ b ≤ c ≤ 1
Chứng minh rằng \(\dfrac{a}{bc+1}+\dfrac{b}{ac+1}+\dfrac{c}{ab+1}\) ≤ 2
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 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 ba số dương 0<a<b<c<1 chứng minh rằng \(\dfrac{a}{bc+1}+\dfrac{b}{ac+1}+\dfrac{c}{ab+1}_-< 2\)
Do a,b,c thuộc N mà a,b,c<1
\(\Rightarrow\)a=0,b=0,c=0
Vậy ....
Cho a,b,c là ba số thực dương thỏa mãn điều kiện ab+bc+ac=3abc. Chứng minh rằng:
\(\sqrt{\dfrac{ab}{a+b+1}}+\sqrt{\dfrac{bc}{b+c+1}}+\sqrt{\dfrac{ca}{c+a+1}}\ge\sqrt{3}\)
cho 3 số thực dương \(0\le a\le b\le c\le1\) .chứng minh rằng \(\dfrac{a}{bc+1}+\dfrac{b}{ac+1}+\dfrac{c}{ab+1}\le2\)
Ta có: \(0\le a\le b\le c\le1\Leftrightarrow\left\{{}\begin{matrix}1-a\ge0\\1-b\ge0\end{matrix}\right.\)
\(\Rightarrow\left(1-a\right)\left(1-b\right)\ge0\Leftrightarrow1\left(1-b\right)-a\left(1-b\right)\ge0\)
\(\Rightarrow1-b-a+ab\ge0\Leftrightarrow1+ab\ge a+b\)
Tiếp tục chứng minh ta có: \(\left\{{}\begin{matrix}1\ge c\\0\le a\le b\Leftrightarrow ab\ge0\end{matrix}\right.\)
cộng theo vế: \(1+ab+1+ab\ge a+b+c+0\)
\(\Rightarrow2\left(1+ab\right)\ge a+b+c\)
Ta có: \(\dfrac{c}{ab+1}=\dfrac{2c}{2\left(ab+1\right)}\le\dfrac{2c}{a+b+c}\) (1)
chứng minh tương tự suy ra đpcm
Ta có: 0≤a≤b≤c≤1⇔{1−a≥01−b≥00≤a≤b≤c≤1⇔{1−a≥01−b≥0
⇒(1−a)(1−b)≥0⇔1(1−b)−a(1−b)≥0⇒(1−a)(1−b)≥0⇔1(1−b)−a(1−b)≥0
⇒1−b−a+ab≥0⇔1+ab≥a+b⇒1−b−a+ab≥0⇔1+ab≥a+b
Tiếp tục chứng minh ta có: {1≥c0≤a≤b⇔ab≥0{1≥c0≤a≤b⇔ab≥0
cộng theo vế: 1+ab+1+ab≥a+b+c+01+ab+1+ab≥a+b+c+0
⇒2(1+ab)≥a+b+c⇒2(1+ab)≥a+b+c
Ta có: cab+1=2c2(ab+1)≤2ca+b+ccab+1=2c2(ab+1)≤2ca+b+c (1)
Cho \(a;b;c\) là các số thực dương thỏa mãn :\(0< a;b;c< 1\). Chứng minh rằng:
\(\dfrac{1}{a.\left(1-b\right)}+\dfrac{1}{b.\left(1-c\right)}+\dfrac{1}{c.\left(1-a\right)}\ge\dfrac{3}{1-\left(a+b+c\right)+ab+bc+ac}\)
P/s: Đề cương toán lớp 10 trường THPT chuyên sư phạm Hà Nội.
Em xin nhờ quý thầy cô giáo và các bạn giúp đỡ, em cám ơn nhiều ạ!
Đặt \(a\left(1-b\right)=x;b\left(1-c\right)=y;c\left(1-a\right)=x\)
\(\Rightarrow1-\left(a+b+c\right)+ab+bc+ca=1-a\left(1-b\right)-b\left(1-c\right)-c\left(1-a\right)=1-x-y-z\)
BĐT cần c/m trở thành:
\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge\dfrac{3}{1-x-y-z}\)
\(\Leftrightarrow\left(1-x-y-z\right)\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)-3\ge0\)
\(\Leftrightarrow\dfrac{1-x-y-z}{x}+\dfrac{1-x-y-z}{y}+\dfrac{1-x-y-z}{z}-3\ge0\)
\(\Leftrightarrow\dfrac{1-y-z}{x}+\dfrac{1-z-x}{y}+\dfrac{1-x-y}{z}-6\ge0\) (1)
Lại có: \(1-y-z=1-b\left(1-c\right)-c\left(1-a\right)=1-b-c+bc+ca=\left(1-b\right)\left(1-c\right)+ca\)
Nên (1) tương đương:
\(\dfrac{\left(1-b\right)\left(1-c\right)+ca}{a\left(1-b\right)}+\dfrac{\left(1-a\right)\left(1-c\right)+ab}{b\left(1-c\right)}+\dfrac{\left(1-a\right)\left(1-b\right)+bc}{c\left(1-a\right)}-6\ge0\)
\(\Leftrightarrow\dfrac{1-c}{a}+\dfrac{c}{1-b}+\dfrac{1-a}{b}+\dfrac{a}{1-c}+\dfrac{1-b}{c}+\dfrac{b}{1-a}\ge6\)
BĐT trên hiển nhiên đúng theo AM-GM do:
\(\dfrac{1-c}{a}+\dfrac{c}{1-b}+\dfrac{1-a}{b}+\dfrac{a}{1-c}+\dfrac{1-b}{c}+\dfrac{b}{1-a}\ge6\sqrt[6]{\dfrac{abc\left(1-a\right)\left(1-b\right)\left(1-c\right)}{abc\left(1-a\right)\left(1-b\right)\left(1-c\right)}}=6\) (đpcm)
Dấu "=" xảy ra khi \(a=b=c=\dfrac{1}{2}\)
Cho a,b,c > 0 và ab + bc + ac = 1. Chứng minh rằng :\(\dfrac{a}{\sqrt{a^2+1}}+\dfrac{b}{\sqrt{b^2+1}}+\dfrac{c}{\sqrt{c^2+1}}\le\dfrac{3}{2}\)
\(\dfrac{a}{\sqrt{a^2+1}}=\dfrac{a}{\sqrt{a^2+ab+ac+bc}}=\dfrac{a}{\sqrt{\left(a+b\right)\left(a+c\right)}}\le\dfrac{a}{2}\left(\dfrac{1}{a+b}+\dfrac{1}{a+c}\right)=\dfrac{1}{2}\left(\dfrac{a}{a+b}+\dfrac{a}{a+c}\right)\) Chứng minh tương tự ta được:
\(\dfrac{b}{\sqrt{b^2+1}}\le\dfrac{1}{2}\left(\dfrac{b}{b+a}+\dfrac{b}{b+c}\right);\dfrac{c}{\sqrt{c^2+1}}\le\dfrac{1}{2}\left(\dfrac{c}{c+a}+\dfrac{c}{c+b}\right)\)
\(\Rightarrow\dfrac{a}{\sqrt{a^2+1}}+\dfrac{b}{\sqrt{b^2+1}}+\dfrac{c}{\sqrt{c^2+1}}\le\dfrac{1}{2}\left(\dfrac{a}{a+b}+\dfrac{a}{a+c}+\dfrac{b}{b+a}+\dfrac{b}{b+c}+\dfrac{c}{c+a}+\dfrac{c}{c+b}\right)=\dfrac{1}{2}\left(\dfrac{a+b}{a+b}+\dfrac{b+c}{b+c}+\dfrac{c+a}{c+a}\right)=\dfrac{1}{2}\left(1+1+1\right)=\dfrac{3}{2}\) Dấu = xảy ra \(\Leftrightarrow a=b=c=\dfrac{1}{\sqrt{3}}\)
\(\dfrac{a}{\sqrt{a^2+1}}=\dfrac{a}{\sqrt{a^2+ab+bc+ca}}=\dfrac{a}{\sqrt{\left(a+b\right)\left(a+c\right)}}\le\dfrac{1}{2}\left(\dfrac{a}{a+b}+\dfrac{a}{a+c}\right)\)
Tương tự: \(\dfrac{b}{\sqrt{b^2+1}}\le\dfrac{1}{2}\left(\dfrac{b}{a+b}+\dfrac{b}{b+c}\right)\) ; \(\dfrac{c}{\sqrt{c^2+1}}\le\dfrac{1}{2}\left(\dfrac{c}{c+a}+\dfrac{c}{b+c}\right)\)
Cộng vế:
\(VT\le\dfrac{1}{2}\left(\dfrac{a}{a+b}+\dfrac{b}{a+b}+\dfrac{a}{a+c}+\dfrac{c}{a+c}+\dfrac{b}{b+c}+\dfrac{c}{b+c}\right)=\dfrac{3}{2}\)
Dấu "=" xảy ra khi \(a=b=c=\dfrac{1}{\sqrt{3}}\)
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