Question

cho \(a+b+c=1\) và  \(a,b,c\ge-\frac{1}{4}\) chứng minh rằng :

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

Answer 1

Ap dung BDT Bun-hia-cop-xki ta co:

\(\left(\sqrt{4a+1}+\sqrt{4b+1}+\sqrt{4c+1}\right)^2\le\left(1+1+1\right)\left[4\left(a+b+c\right)+3\right]=21\)

\(\Rightarrow-\sqrt{21}\le\sqrt{4a+1}+\sqrt{4b+1}+\sqrt{4c+1}\le\sqrt{21}< 5\)

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

Answer 2

Ap dung BDT Bun-hia-cop-xki ta co:

\left(\sqrt{4a+1}+\sqrt{4b+1}+\sqrt{4c+1}\right)^2\le\left(1+1+1\right)\left[4\left(a+b+c\right)+3\right]=21(4a+1​+4b+1​+4c+1​)2≤(1+1+1)[4(a+b+c)+3]=21

\Rightarrow-\sqrt{21}\le\sqrt{4a+1}+\sqrt{4b+1}+\sqrt{4c+1}\le\sqrt{21}&lt; 5⇒−21​≤4a+1​+4b+1​+4c+1​≤21​<5

\Rightarrow\sqrt{4a+1}+\sqrt{4b+1}+\sqrt{4c+1}&lt; 5⇒4a+1​+4b+1​+4c+1​<5