\(\dfrac{9ab}{ab+a+b}\)\(\le\)1+a+b
\(\Rightarrow\)9ab\(\le\)(1+a+b)(a+b+ab)
Xét 9ab = [3\(\sqrt{ab}\)]\(^2\) = [\(\sqrt{ab}\) + \(\sqrt{ab}\) + \(\sqrt{ab}\)]\(^2\) = [1.\(\sqrt{ab}\) + \(\sqrt{a}\).\(\sqrt{b}\)+ \(\sqrt{b}\).\(\sqrt{a}\)]\(^2\) ≤ (1\(^2\) + a + b)( ab + b + a)
Dấu = xảy ra khi a = b = 1
1. Tìm GTLN \(y=x^3\left(2-x\right)^5\)
2. Cho \(0\le a\le1\). Chứng minh rằng \(a\left(1-a^2\right)\)\(\le\dfrac{2}{3\sqrt{3}}\)
3. Cho a,b,c >0
CMR: \(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\ge\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\)
Cho tam giác ABC có BC=a,AC=b,AB=c. Chứng minh rằng: \(3\left(a^3+b^3+c^3\right)+4abc\ge\dfrac{13}{27}\left(a+b+c\right)^3\)
Cho a,b,c > 0 và các số x,y,z dương . CHứng minh rằng
\(\dfrac{a\left(z^2+y^2\right)}{b+c}+\dfrac{b\left(x^2+z^2\right)}{a+c}+\dfrac{c\left(x^2+y^2\right)}{a+b}\ge xy+yz+xz\)
Cho a,b,c >0 .Chứng minh:
\(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}+\dfrac{2\left(ab+bc+ca\right)}{a^2+b^2+c^2}\ge5\)
cho 3 số a, b, c > 0 thỏa mãn a + b + c = 1. Chứng minh
\(\dfrac{a}{a+\sqrt{2019a+bc}}+\dfrac{b}{b+\sqrt{2019b+ac}}+\dfrac{c}{c+\sqrt{2019c+ab}}\le1\)
Cho a,b,c>0.Chứng minh rằng:\(\dfrac{a^2+b^2}{a+b}+\dfrac{b^2+c^2}{b+c}+\dfrac{c^2+a^2}{a+c}\le\dfrac{3.\left(a^2+b^2+c^2\right)}{a+b+c}\)
Cho a , b , c là các số thực dương . Chứng minh rằng
\(\dfrac{b^2c}{a^3\left(b+c\right)}+\dfrac{c^2a}{b^3\left(c+a\right)}+\dfrac{a^2b}{c^3\left(a+b\right)}\ge\dfrac{1}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
Cho a,b,c dương. Chứng minh
\(\dfrac{1}{\left(a+b\right)^2}+\dfrac{1}{\left(b+c\right)^2}+\dfrac{1}{\left(c+a\right)^2}\ge\dfrac{3\sqrt{3abc\left(a+b+c\right)}.\left(a+b+c\right)^2}{4\left(ab+bc+ca\right)^3}\)
Với mọi số thực dương a,b,c. chứng minh rằng:
4(\(\dfrac{a^2b}{c}+\dfrac{b^2c}{a}+\dfrac{c^2a}{b}\))+8\(\left(\dfrac{c}{\left(2a+b\right)^2}+\dfrac{b}{\left(2c+a\right)^2}+\dfrac{a}{\left(2b+c\right)^2}\right)\ge3\left(a+b+c\right)\)
