\(\frac{a}{\sqrt{b}-1}+4(\sqrt{b}-1)\ge 2\sqrt{\frac{a}{\sqrt{b}-1}\cdot 4(\sqrt{b}-1)}=2\sqrt{4a}=4\sqrt{a}\). Tương tự, ta có: \(\frac{b}{\sqrt{c}-1}+4(\sqrt{c}-1)\ge 4\sqrt{b}\). \(\frac{c}{\sqrt{a}-1}+4(\sqrt{a}-1)\ge 4\sqrt{c}\).  Cộng các bất đẳng thức: Cộng ba bất đẳng thức trên, ta được: \(\frac{a}{\sqrt{b}-1}+\frac{b}{\sqrt{c}-1}+\frac{c}{\sqrt{a}-1}+4(\sqrt{b}-1)+4(\sqrt{c}-1)+4(\sqrt{a}-1)\ge 4(\sqrt{a}+\sqrt{b}+\sqrt{c})\). Sắp xếp lại, ta có: \(\frac{a}{\sqrt{b}-1}+\frac{b}{\sqrt{c}-1}+\frac{c}{\sqrt{a}-1}\ge 4(\sqrt{a}+\sqrt{b}+\sqrt{c})-4(\sqrt{a}-1)-4(\sqrt{b}-1)-4(\sqrt{c}-1)\). \(\frac{a}{\sqrt{b}-1}+\frac{b}{\sqrt{c}-1}+\frac{c}{\sqrt{a}-1}\ge 4\sqrt{a}+4\sqrt{b}+4\sqrt{c}-4\sqrt{a}+4-4\sqrt{b}+4-4\sqrt{c}+4\). \(\frac{a}{\sqrt{b}-1}+\frac{b}{\sqrt{c}-1}+\frac{c}{\sqrt{a}-1}\ge 12\).

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)

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

ai làm đi

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