\(\left(\sqrt{a}+\sqrt{b}\right)^8=\left[\left(\sqrt{a}+\sqrt{b}\right)^2\right]^4=\left[a+b+2\sqrt{ab}\right]^4\)
áp dụng BDT AM-GM
\(=>\left[a+b+2\sqrt{ab}\right]^4\ge\left[2\sqrt{\left(a+b\right)\left(2\sqrt{ab}\right)}\right]^4=64ab\left(a+b\right)^2\)
CM \(\left(\sqrt{a}+\sqrt{b}\right)^8\ge64ab\left(a+b\right)^2\) Với \(a,b\ge0\)
CM \(\left(\sqrt{a}+\sqrt{b}\right)^8\ge64ab\left(a+b\right)^2\) Với \(a,b\ge0\)
CMR \(\left(\sqrt{a}+\sqrt{b}\right)^8\ge64ab\left(a+b\right)^2\Leftrightarrow a,b\ge0\)
Chứng minh giúp mình mấy câu bất đẳng thức này nha
a) \(\frac{2\sqrt{ab}}{\sqrt{a}+\sqrt{b}}\le\sqrt[4]{ab}\left(a,b>0\right)\)
b) \(\left(\sqrt{a}+\sqrt{b}\right)^8\ge64ab\left(a+b\right)^2\left(a,b>0\right)\)
c) \(y\left(\frac{1}{x}+\frac{1}{x}\right)+\frac{1}{y}\left(x+z\right)\le\left(\frac{1}{x}+\frac{1}{z}\right)\left(x+z\right)\left(0< x\le y\le z\right)\)
d) \(a+b+c\ge3\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\left(a,b,c>0;a+b+c=abc\right)\)
Ta có \(\sqrt{a+b}+\sqrt{a-b}\le2\sqrt{a}\)
\(\Leftrightarrow\left(\sqrt{a+b}+\sqrt{a-b}\right)^2\le\left(2\sqrt{a}\right)^2\)\(\Leftrightarrow a+b+a-b+2.\sqrt{\left(a+b\right)\left(a-b\right)}\le4a\)
\(\Leftrightarrow2a+2\sqrt{\left(a+b\right)\left(a-b\right)}\le4a\)
\(\Leftrightarrow-2a+2.\sqrt{\left(a+b\right)\left(a-b\right)}\le0\)\(\Leftrightarrow-\left(2a-2.\sqrt{\left(a+b\right).\left(a-b\right)}\right)\le0\)
\(\Leftrightarrow a+b+a-b-2.\sqrt{\left(a+b\right)\left(a-b\right)}\ge0\)
\(\Leftrightarrow\left(\sqrt{a+b}-\sqrt{a-b}\right)^2\ge0\)( luôn đúng nên suy ra điều phải chứng minh )
1. Cho \(a\ge0;b\ge0.\) Chứng minh: \(\left(a+1\right)\sqrt{b}+\left(b+1\right)\sqrt{a}\le\left(a+1\right)\left(b+1\right)\)
2. Cho \(a\ge2;b\ge\frac{1}{2}\)Chứng minh: \(a\sqrt{2b-1}+2b\sqrt{2a-4}\le2ab\)
Rút gọn biểu thức
a) \(\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2+4\sqrt{ab}}{\left(\sqrt{a+\sqrt{b}}\right)^2-4\sqrt{ab}}.\dfrac{a-b}{\left(\sqrt{a}-\sqrt{b}\right)^2}\) \(\left(đkxđ:a\ne b;a\ge0;b\ge0\right)\)
b) \(\dfrac{a+b-2\sqrt{ab}}{\sqrt{a}-\sqrt{b}}-\dfrac{a-b}{\left(\sqrt{a}+\sqrt{b}\right)^2}\)\(\left(đkxđ:a\ne b;a\ge0;b\ge0\right)\)
HELP ME PLSSSSSSSSSS
Cho \(a,b,c\ge0\)Thỏa mãn: a + b + c = 1010
Chứng minh: \(\sqrt{2020a+\frac{\left(b-c\right)^2}{2}}+\sqrt{2020b+\frac{\left(c-a\right)^2}{2}}+\sqrt{2020c+\frac{\left(a-b\right)^2}{2}}\le2020\sqrt{2}\)
Chứng minh các đẳng thức sau
a) \(\left(\frac{2\sqrt{6}-\sqrt{3}}{2\sqrt{2}-1}+\frac{5+2\sqrt{5}}{2+\sqrt{5}}\right)\left(\sqrt{5}-\sqrt{3}\right)\)
b) \(\frac{a-b}{b^2}\sqrt{\frac{a^2b^4}{a^2-2ab+b^2}}=-a\)(Với b<a<0
c)\(\left(\sqrt{a}+\frac{1-a\sqrt{a}}{1-\sqrt{a}}\right)\left(\frac{1-\sqrt{a}}{1-a}\right)^2=1\)với a\(\ge0\),a khác 1
d) \(\left(\frac{3\sqrt{5}-\sqrt{15}}{\sqrt{27}-3}+\frac{2\sqrt{5}}{\sqrt{3}}\right)40\sqrt{15}=600\)
e) \(\left(1+\frac{x+\sqrt{x}}{\sqrt{x}+1}\right)\left(1-\frac{x-\sqrt{x}}{\sqrt{x}-1}\right)=1-x\)với x\(\ge0;x\ne1\)
