Question

[đỗ mạnh quang]

cho a,b,c>0 và a+b+c=1 cmr căn(4a+1)+căn(4b+1)+căn(4c+1)<5

Answer 1

[Đinh Đức Hùng]

Áp dụng bđt Cauchy ta có :

\(\sqrt{4a+1}\le\frac{4a+1+1}{2}=2a+1\)

\(\sqrt{4b+1}\le\frac{4b+1+1}{2}=2b+1\)

\(\sqrt{4c+1}\le\frac{4c+1+1}{2}=2c+1\)

\(\Rightarrow\sqrt{4a+1}+\sqrt{4b+1}+\sqrt{4b+1}\le2\left(a+b+c\right)+3=5\)(đpcm)

Answer 2

[pham trung thanh]

Áp dụng BĐT Bu-nhi-a-cốp-ski, ta có: 

\(\left(1+1+1\right)\left[\left(\sqrt{4a+1}\right)^2+\left(\sqrt{4b+1}\right)^2+\left(\sqrt{4c+1}\right)^2\right]\)

\(\ge\left(\sqrt{4a+1}+\sqrt{4b+1}+\sqrt{4c+1}\right)^2\)

\(\Leftrightarrow\left(\sqrt{4a+1}+\sqrt{4b+1}+\sqrt{4c+1}\right)^2\le3\left(4a+1+4b+1+4c+1\right)\)

\(\Leftrightarrow VT^2\le21\)

\(\Rightarrow VT^2< 25\)

\(\Rightarrow VT< 5\)

Vậy \(\sqrt{4a+1}+\sqrt{4c+1}+\sqrt{4b+1}< 5\)