cho a,b,c là các số ko âm và a+b+c=1 . Chứng minh : $\sqrt[]{a+1}$+ $\sqrt[]{b+1}$ +$\sqrt[]{c+1}$ < 3,5

Giải thích các bước giải:

 $\begin{cases}\sqrt{a+1}=\dfrac{\sqrt{3}}{2}\sqrt{\dfrac{4}{3}(a+1)}= \dfrac{\sqrt{3}}{4}.2\sqrt{\dfrac{4}{3}(a+1)}\le \dfrac{\sqrt{3}}{4}.(\dfrac{4}{3}+a+1)=\dfrac{\sqrt{3}}{4}.(\dfrac{7}{3}+a)\\\sqrt{b+1}=\dfrac{\sqrt{3}}{2}\sqrt{\dfrac{4}{3}(b+1)}= \dfrac{\sqrt{3}}{4}.2\sqrt{\dfrac{4}{3}(b+1)}\le \dfrac{\sqrt{3}}{4}.(\dfrac{4}{3}+b+1)=\dfrac{\sqrt{3}}{4}.(\dfrac{7}{3}+b)\\\sqrt{c+1}=\dfrac{\sqrt{3}}{2}\sqrt{\dfrac{4}{3}(c+1)}= \dfrac{\sqrt{3}}{4}.2\sqrt{\dfrac{4}{3}(c+1)}\le \dfrac{\sqrt{3}}{4}.(\dfrac{4}{3}+c+1)=\dfrac{\sqrt{3}}{4}.(\dfrac{7}{3}+c)\end{cases}$

$\rightarrow\sqrt{a+1}+\sqrt{b+1}+\sqrt{c+1}\le \dfrac{\sqrt{3}}{4}.(\dfrac{7}{3}.3+a+b+c)=2\sqrt{3}<3,5$