cho a,b,c là các số thục không âm . CMR :
\(a\sqrt{4a^2+5bc}+b\sqrt{4b^2+5ca}+c\sqrt{4c^2+5ab}\ge\left(a+b+c\right)^2\)
Cho a,b,c>0. CMR: \(\sqrt[3]{\frac{a^2}{\left(b+c\right)^2+5bc}}+\sqrt[3]{\frac{b^2}{\left(c+a\right)^2+5ca}}+\sqrt[3]{\frac{c^2}{\left(a+b\right)^2+5ab}}\ge\sqrt[3]{3}\)
cho a;b;c là các số thực dương thỏa mãn \(a^2+b^2+c^2=\frac{1}{3}\)CMR:\(\sqrt{\frac{\left(a+b\right)^3}{8ab\left(4a+4b+c\right)}}+\sqrt{\frac{\left(b+c\right)^3}{8bc\left(4b+4c+a\right)}}+\sqrt{\frac{\left(c+a\right)^3}{8ca\left(4c+4a+b\right)}}\ge a+b+c\)
cho a,b,c >0. chứng minh \(\frac{a}{\sqrt{4a^2+5bc}}+\frac{b}{\sqrt{4b^2+5ac}}+\frac{c}{\sqrt{4c^2+5ab}}\le1.\)
Cho a. b. c không âm và có tổng bằng 1. Chứng minh rằng:
\(\frac{a}{4b^2+1}+\frac{b}{4c^2+1}+\frac{c}{4a^2+1}\ge\left(a\sqrt{a}+b\sqrt{b}+c\sqrt{c}\right)^2\)
đặt \(S=\frac{a}{4b^2+1}+\frac{b}{4c^2+1}+\frac{c}{4a^2+1}\)
\(=\frac{a^3}{4a^2b^2+a^2}+\frac{b^3}{4b^2c^2+b^2}+\frac{c^3}{4a^2c^2+c^2}\ge\frac{\left(a\sqrt{a}+b\sqrt{b}+c\sqrt{c}\right)^2}{4a^2b^2+4b^2c^2+4c^2a^2+a^2+b^2+c^2}\)
xét hiệu:
1-4(a2b2+b2c2+c2a2)-a2-b2-c2
=2ab+2bc+2ca-4(a2b2+b2c2+c2a2)
=2ab(1-2ab)+2bc(1-2bc)+2ca(1-2ca)
ta có:
\(2ab\le\frac{\left(a+b\right)^2}{2}\le\frac{1}{2};2bc\le\frac{\left(b+c\right)^2}{2}\le\frac{1}{2};2ca\le\frac{\left(c+a\right)^2}{2}\le\frac{1}{2}\)
\(\Rightarrow2ab\left(1-2ab\right);2bc\left(1-2bc\right);2ca\left(1-2ca\right)\ge0\)
\(\Rightarrow1\ge4\left(a^2b^2+b^2c^2+c^2a^2\right)+a^2+b^2+c^2\)
\(\Rightarrow\frac{\left(a\sqrt{a}+b\sqrt{b}+c\sqrt{c}\right)^2}{4\left(a^2b^2+b^2c^2+c^2a^2\right)+a^2+b^2+c^2}\ge\left(a\sqrt{a}+b\sqrt{b}+c\sqrt{c}\right)^2\)
\(\Rightarrow\frac{a}{4b^2+1}+\frac{b}{4c^2+1}+\frac{c}{4a^2+1}\ge\left(a\sqrt{a}+b\sqrt{b}+c\sqrt{c}\right)^2\)
=>đpcm
dấu"=" xảy ra khi 1 số=1;2 số còn lại =0
Cho a,b,c thực dương .CMR
\(\sqrt{\frac{\left(a+b\right)^3}{ab\left(4a+4b+c\right)}}+\sqrt{\frac{\left(b+c\right)^3}{bc\left(4b+4c+a\right)}}+\sqrt{\frac{\left(c+a\right)^3}{ca\left(4c+4c+b\right)}}\ge2\sqrt{2}\)
Gọi A là vế trái của BĐT cần chứng minh. Không mất tính tổng quát, ta giả sử a + b + c = 3. Áp dụng BĐT AM - GM ta có:
\(\sqrt{\frac{\left(a+b\right)^3}{8ab\left(4a+4b+c\right)}}+\sqrt{\frac{\left(a+b\right)^3}{8bc\left(4a+4b+c\right)}}+\frac{ab\left(4a+4b+c\right)}{27}\)\(\ge\frac{1}{2}\left(a+b\right)\)
Suy ra
\(\sqrt{\frac{\left(a+b\right)^3}{8ab\left(4a+4b+c\right)}}\)\(+\frac{ab\left(4a+4b+c\right)}{54}\ge\frac{1}{4}\left(a+b\right)\)
Tương tự
\(\sqrt{\frac{\left(b+c\right)^3}{8bc\left(4b+4c+a\right)}}+\frac{bc\left(4b+4c+a\right)}{54}\ge\frac{1}{4}\left(b+c\right)\)
và \(\sqrt{\frac{\left(c+a\right)^3}{8ca\left(4c+4a+b\right)}}+\frac{ca\left(4c+4a+b\right)}{54}\ge\frac{1}{4}\left(c+a\right)\)
Cộng ba BĐT trên ta có:
\(\frac{1}{2\sqrt{2}}A\ge B\)
Với \(A=\frac{1}{54}[ab\left(4a+4b+c\right)+bc\left(4b+4c+a\right)\)
\(+ca\left(4c+4a+b\right)]\)
\(=\frac{1}{54}\left[4ab\left(a+b\right)+4bc\left(b+c\right)+4ca\left(c+a\right)+3abc\right]\)
\(=\frac{1}{54}\left[4\left(a+b+c\right)\left(ab+bc+ca\right)-9abc\right]\)
\(\le\frac{1}{54}\left(a+b+c\right)^3=\frac{1}{2}\)
và \(B=\frac{1}{4}.2\left(a+b+c\right)=\frac{3}{2}\)
Suy ra \(\frac{1}{2\sqrt{2}}A\ge\frac{3}{2}-\frac{1}{2}=1\Rightarrow A\ge2\sqrt{2}\)
Vậy
\(\sqrt{\frac{\left(a+b\right)^3}{ab\left(4a+4b+c\right)}}+\sqrt{\frac{\left(a+b\right)^3}{bc\left(4a+4b+c\right)}}+\sqrt{\frac{\left(c+a\right)^3}{ca\left(4c+4a+b\right)}}\ge2\sqrt{2}\)(đpcm)
toán lớp 5 phiên bản hack não
Cho a,b,c>0 CMR
\(\sqrt{\dfrac{a^2}{6a^2+5ab+b^2}}+\sqrt{\dfrac{b^2}{6b^2+5bc+c^2}}+\sqrt{\dfrac{c^2}{6c^2+5ca+a^2}}\le\dfrac{\sqrt{3}}{2}\)
Với các số không âm a, b, c sao cho không có 2 số nào đồng thời bằng 0 và a+ b+ c= 2. CMR:
\(\frac{a}{\sqrt{4a+3bc}}+\frac{b}{\sqrt{4b+3ca}}+\frac{c}{\sqrt{4c+3ab}}\le1\)
cho a,b,c là các số không âm. CMR
a+b+c\(\ge\)\(\sqrt{ab}\)+\(\sqrt{bc}\)+\(\sqrt{ca}\)+ \(\frac{\left(\sqrt{a}-\sqrt{b}\right)^2+\left(\sqrt{b}-\sqrt{c}\right)^2+\left(\sqrt{c}-\sqrt{a}\right)^2}{2016}\)
Cho a, b, c là các số thực dương. Chứng minh rằng :
\(\frac{a}{\sqrt{4a^2+\left(b+c\right)^2}}+\frac{b}{\sqrt{4b^2+\left(c+a\right)^2}}+\frac{c}{\sqrt{4c^2+\left(a+b\right)^2}}\le\frac{3\sqrt{2}}{4}\)
\(P=\sum\frac{a}{\sqrt{\left(2a\right)^2+\left(b+c\right)^2}}\le\sqrt{2}\sum\frac{a}{2a+b+c}=\sqrt{2}\sum a\left(\frac{1}{a+b+a+c}\right)\le\frac{\sqrt{2}}{4}\sum\left(\frac{a}{a+b}+\frac{a}{a+c}\right)=\frac{3\sqrt{2}}{4}\)
Dấu "=" xảy ra khi \(a=b=c\)