ta có:
\(3\left(a^4+b^4+c^4\right)\ge\left(a^2+b^2+c^2\right)^2\ge\left(ab^2+bc^2+ca^2\right)\left(a+b+c\right)\)
\(\Rightarrow a^4+b^4+c^4\ge a+b+c\)
lại có:
\(\left(a^4+b^4+c^4\right)\left(b^2+a^2+c^2\right)\ge\left(ab^2+bc^2+ca^2\right)^2=9\)
\(\Rightarrow\left(a^4+b^4+c^4\right).\sqrt{3\left(a^4+b^4+c^4\right)}\ge9\)
\(\Rightarrow a^4+b^4+c^4\ge3\)
\(\Rightarrow24\left(a^4+b^4+c^4\right)\ge a+b+c+69\ge12\sqrt[3]{a+7}+...\)
\(P=\frac{a}{\sqrt{\left(b+1\right)\left(b^2-b+1\right)}}+\frac{b}{\sqrt{\left(c+1\right)\left(c^2-c+1\right)}}+\frac{c}{\sqrt{\left(a+1\right)\left(a^2-a+1\right)}}\)
\(\ge\frac{2a}{b^2+2}+\frac{2b}{c^2+2}+\frac{2c}{a^2+2}=\left(a+b+c\right)-\left(\frac{ab^2}{b^2+2}+\frac{bc^2}{c^2+2}+\frac{ca^2}{a^2+2}\right)\)
\(=6-\left(\frac{2ab^2}{b^2+4+b^2}+\frac{2bc^2}{c^2+4+c^2}+\frac{2ca^2}{a^2+4+a^2}\right)\ge6-\left(\frac{2ab}{b+4}+\frac{2bc}{c+4}+\frac{2ca}{a+4}\right)\)
\(=6-\left(2a+2b+2c-\frac{8a}{b+4}-\frac{8b}{c+4}-\frac{8c}{a+4}\right)\)
\(=\frac{8a}{b+4}+\frac{8b}{c+4}+\frac{8c}{a+4}-6=\frac{8a^2}{ab+4a}+\frac{8b^2}{bc+4b}+\frac{8c^2}{ca+4c}-6\)
\(\ge\frac{8\left(a+b+c\right)^2}{\left(ab+bc+ca\right)+4\left(a+b+c\right)}-6\ge\frac{288}{\frac{\left(a+b+c\right)^2}{3}+24}-6=2\)
Bài 1 :Cho a,b,c dương thỏa mãn a+b+c=2
CMR \(\frac{bc}{\sqrt{3a^2+4}}+\frac{ca}{\sqrt{3b^2+4}}+\frac{ab}{\sqrt{3c^2+4}}\ge\frac{\sqrt{3}}{3}\)
Bài 2:Cho a,b,c>0. CMR
\(\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge\frac{8}{9}\left(a+b+c\right)\left(ab+bc+ca\right)\)
Cho các số thực dương a, b, c. Chứng minh rằng:
\(\sqrt{c^2\left(a^2+b^2\right)^2+a^2\left(b^2+c^2\right)^2+b^2\left(c^2+a^2\right)^2}\ge\frac{54\left(abc\right)^3}{\left(a+b+c\right)^2\sqrt{\left(ab\right)^4+\left(bc\right)^4+\left(ca\right)^4}}\)
áp dụng cô si ta có:
+)\(\frac{a^5}{b^3}+\frac{a^3}{b}\ge\frac{2a^4}{b^2};\frac{b^5}{c^3}+\frac{b^3}{c}\ge\frac{2b^4}{c^2};\frac{c^5}{a^3}+\frac{c^3}{a}\ge\frac{2c^4}{a^2}\)
\(\Leftrightarrow\frac{a^5}{b^3}+\frac{b^5}{c^3}+\frac{c^5}{a^3}\ge2\left(\frac{a^4}{b^2}+\frac{b^4}{c^2}+\frac{c^4}{a^2}\right)-\left(\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}\right)\)
+)\(\frac{a^4}{b^2}+a^2\ge\frac{2a^3}{b};\frac{b^4}{c^2}+b^2\ge\frac{2b^3}{c};\frac{c^4}{a^2}+c^2\ge\frac{2C^3}{a}\)
\(\Leftrightarrow\frac{a^4}{b^2}+\frac{b^4}{c^2}+\frac{c^4}{a^2}\ge2\left(\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}\right)-\left(a^2+b^2+c^2\right)\)
+)\(\frac{a^3}{b}+ab\ge2a^2;\frac{b^3}{c}+bc\ge2b^2;\frac{c^3}{a}+ca\ge2c^2\)
\(\Leftrightarrow\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}\ge\left(a^2+b^2+c^2\right)+\left(a^2+b^2+c^2-ab-bc-ca\right)\ge\left(a^2+b^2+c^2\right)\)
\(\Leftrightarrow\frac{a^4}{b^2}+\frac{b^4}{c^2}+\frac{c^4}{a^2}\ge\left(\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}\right)+\left(\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}-a^2-b^2-c^2\right)\ge\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}\)
\(\Leftrightarrow\frac{a^5}{b^3}+\frac{b^5}{c^3}+\frac{c^5}{a^3}\ge\left(\frac{a^4}{b^2}+\frac{b^4}{c^2}+\frac{c^4}{a^2}\right)+\left(\frac{a^4}{b^2}+\frac{b^4}{c^2}+\frac{c^4}{a^2}-\frac{a^3}{b}-\frac{b^3}{c}-\frac{c^3}{a}\right)\ge\left(\frac{a^4}{b^2}+\frac{b^4}{c^2}+\frac{c^4}{a^2}\right)\)
1/Cho các số thực dương chứng minh:\(\frac{3\left(a^4+b^4+c^4\right)}{\left(a^2+b^2+c^2\right)^2}+\frac{ab+bc+ca}{a^2+b^2+c^2}\ge2\)
2/Cho a,b dương.Chứng minh:\(\left(\frac{a}{b}+\frac{b}{a}\right)+4\sqrt{2}\frac{a+b}{\sqrt{a^2+b^2}}\ge10\)
3/ Cho các số thực dương. Chứng minh: \(\left(a^2+2bc\right)\left(b^2+2ca\right)\left(c^2+2ab\right)\ge abc\left(a+2b\right)\left(b+2c\right)\left(c+2a\right)\)
Bài 1:Cho a,b,c,d là các số dương. Chứng minh rằng :
\(\frac{a^4}{\left(a+b\right)\left(a^2+b^2\right)}+\frac{b^4}{\left(b+c\right)\left(b^2+c^2\right)}+\frac{c^4}{\left(c+d\right)\left(c^2+d^2\right)}+\frac{d^4}{\left(d+a\right)\left(d^2+a^2\right)}\ge\frac{a+b+c+d}{4}\)
Bài 2:Cho \(a>0,b>0,c>0\).\(CM:\frac{a}{bc}+\frac{b}{ca}+\frac{c}{ab}\ge2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
Bài 3: a) Cho x,y,>0. CMR:\(\frac{x^3}{x^2+xy+y^2}\ge\frac{2x-y}{3}\)
b) Chứng minh rằng\(\Sigma\frac{a^3}{a^2+ab+b^2}\ge\frac{a+b+c}{3}\)
Câu 19 , Đăk Lắk
Cho các số thực dương x ; y ; z thỏa mãn \(x+2y+3z=2\)
Tìm \(S_{max}=\sqrt{\frac{xy}{xy+3z}}+\sqrt{\frac{3yz}{3yz+x}}+\sqrt{\frac{3xz}{3xz+4y}}\)
Giải
Đặt \(\hept{\begin{cases}x=a\\2y=b\\3z=c\end{cases}}\left(a;b;c\right)>0\Rightarrow a+b+c=2\)
Khi đó \(S=\sqrt{\frac{a.\frac{b}{2}}{a.\frac{b}{2}+c}}+\sqrt{\frac{\frac{b}{2}.c}{\frac{b}{2}.c+a}}+\sqrt{\frac{a.c}{a.c+2b}}\)
\(=\sqrt{\frac{ab}{ab+2c}}+\sqrt{\frac{bc}{bc+2a}}+\sqrt{\frac{ac}{ac+2b}}\)
\(=\sqrt{\frac{ab}{ab+\left(a+b+c\right)c}}+\sqrt{\frac{bc}{bc+\left(a+b+c\right)a}}+\sqrt{\frac{ac}{ac+\left(a+b+c\right)b}}\)
\(=\sqrt{\frac{ab}{ab+ac+bc+c^2}}+\sqrt{\frac{bc}{bc+a^2+ab+ac}}+\sqrt{\frac{ac}{ac+ab+b^2+bc}}\)
\(=\sqrt{\frac{ab}{\left(a+c\right)\left(b+c\right)}}+\sqrt{\frac{bc}{\left(a+b\right)\left(a+c\right)}}+\sqrt{\frac{ac}{\left(a+b\right)\left(b+c\right)}}\)
\(\le\frac{\frac{a}{a+c}+\frac{b}{b+c}}{2}+\frac{\frac{b}{a+b}+\frac{c}{a+c}}{2}+\frac{\frac{a}{a+b}+\frac{c}{b+c}}{2}\left(Cauchy\right)\)
\(=\frac{1}{2}\left(\frac{a}{a+c}+\frac{c}{a+c}\right)+\frac{1}{2}\left(\frac{b}{b+c}+\frac{c}{b+c}\right)+\frac{1}{2}\left(\frac{a}{a+b}+\frac{b}{a+b}\right)\)
\(=\frac{1}{2}+\frac{1}{2}+\frac{1}{2}=\frac{3}{2}\)
Dấu "=" tại a = b = c
20, Thanh hóa
Cho a;b;c > 0 thỏa abc = 1
CMR \(\frac{ab}{a^4+b^4+ab}+\frac{bc}{b^4+c^4+bc}+\frac{ac}{a^4+c^4+ac}\le1\)
Giải
Áp dụng bất đẳng thức Bunhiacopxki có
\(\left(a^2+b^2\right)^2\le\left(1+1\right)\left(a^4+b^4\right)\)
\(\Rightarrow a^4+b^4\ge\frac{\left(a^2+b^2\right)^2}{2}=\frac{\left(a^2+b^2\right)\left(a^2+b^2\right)}{2}\ge\frac{2ab\left(a^2+b^2\right)}{2}=ab\left(a^2+b^2\right)\)
\(\Rightarrow a^4+b^4\ge ab\left(a^2+b^2\right)\)
Khi đó \(\frac{ab}{a^4+b^4+ab}\le\frac{ab}{ab\left(a^2+b^2\right)+ab}=\frac{1}{a^2+b^2+1}\)
Chứng minh tương tự \(\frac{bc}{b^4+c^4+bc}\le\frac{1}{b^2+c^2+1}\)
\(\frac{ac}{a^4+c^4+ac}\le\frac{1}{a^2+c^2+1}\)
Khi đó \(VT\le\frac{1}{a^2+b^2+1}+\frac{1}{b^2+c^2+1}+\frac{1}{a^2+c^2+1}=A\)
Ta sẽ chứng minh A < 1
Thật vậy
Đặt \(\left(a^2;b^2;c^2\right)\rightarrow\left(x^3;y^3;z^3\right)\)
\(\Rightarrow xyz=1\)
Khi đó \(A=\frac{1}{x^3+y^3+1}+\frac{1}{y^3+z^3+1}+\frac{1}{z^3+x^3+1}\)
Áp dụng bđt Cô-si có \(x^3+y^3=\left(x+y\right)\left(x^2+y^2-xy\right)\ge\left(x+y\right)\left(2xy-xy\right)=xy\left(x+y\right)\)
\(\Rightarrow x^3+y^3\ge\left(x+y\right)xy\)
\(\Rightarrow x^3+y^3+1\ge\left(x+y\right)xy+1=\left(x+y\right)xy+xyz=xy\left(x+y+z\right)\)
\(\Rightarrow\frac{1}{x^3+y^3+1}\le\frac{1}{xy\left(x+y+z\right)}=\frac{xyz}{xy\left(x+y+z\right)}=\frac{z}{x+y+z}\)
Chứng minh tương tự \(\frac{1}{y^3+z^3+1}\le\frac{x}{x+y+z}\)
\(\frac{1}{x^3+z^3+1}\le\frac{z}{x+y+z}\)
Khi đó \(A\le\frac{x+y+z}{x+y+z}=1\left(đpcm\right)\)
Dấu "=" tại x = y = z = 1
Đang trong quá trình cập nhật những câu tiếp theo , những câu tiếp theo sẽ ở trong phần bình luận
\(\Leftrightarrow\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}+\frac{ab}{a+b}+\frac{bc}{b+c}+\frac{ca}{c+a}\ge a+b+c\)
\(\Leftrightarrow\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}+a-\frac{a^2}{a+b}+b-\frac{b^2}{b+c}+c-\frac{c^2}{c+a}\ge a+b+c\)
\(\Leftrightarrow\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\ge\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+a}\)
\(\Leftrightarrow a^2\left(a+b\right)\left(a+c\right)+b^2\left(b+a\right)\left(b+c\right)+c^2\left(c+a\right)\left(c+b\right)\ge a^2\left(a+c\right)\left(b+c\right)+b^2\left(b+a\right)\left(c+a\right)+c^2\left(c+b\right)\left(a+b\right)\)
\(\Leftrightarrow a^4+b^4+c^4\ge a^2c^2+a^2b^2+b^2c^2\left(lđ\right)\)
\(\Leftrightarrow\frac{a^2+bc}{b+c}+\frac{b^2+ca}{c+a}+\frac{c^2+ab}{a+b}\ge a+b+c\)
Cho a,b,c > 0 thỏa mãn a + b + c = 3.
Chứng minh rằng: \(\frac{a^4}{\left(b+c\right)\left(b^2+c^2\right)}+\frac{b^4}{\left(c+a\right)\left(c^2+a^2\right)}+\frac{c^4}{\left(a+b\right)\left(a^2+b^2\right)}\ge\frac{3}{4}\)
