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

Sử dụng bđt cô-si cho 3 số là ok

\(a^4b^4+b^4c^4+c^4a^4\ge3\sqrt[3]{a^4b^4b^4c^4c^4a^4}=3a^4b^4c^4\)

P/S: Cái gt hơi thừa thì phải ???

Ấy chết pẹ , nhầm , bài nãy sai bỏ đi nha

Tay nhanh hơn não :)) nếu dễ thì t đâu có hỏi ?

hgggggg

Ta có:

\(\left(\sqrt{a}.\dfrac{\sqrt{a}}{\sqrt{4a+3bc}}+\sqrt{b}\dfrac{\sqrt{b}}{\sqrt{4b+3ac}}+\sqrt{c}\dfrac{\sqrt{c}}{\sqrt{4c+3ab}}\right)^2\le\left(a+b+c\right)\left(\dfrac{a}{4a+3bc}+\dfrac{b}{4b+3ac}+\dfrac{c}{4c+3ab}\right)\)

\(=2\left(\dfrac{a}{4a+3bc}+\dfrac{b}{4b+3ac}+\dfrac{c}{4c+3ab}\right)\)

Nên ta chỉ cần chứng minh:

\(\dfrac{a}{4a+3bc}+\dfrac{b}{4b+3ac}+\dfrac{c}{4c+3ab}\le\dfrac{1}{2}\)

\(\Leftrightarrow\dfrac{4a}{4a+3bc}+\dfrac{4b}{4b+3ac}+\dfrac{4c}{4c+3ab}\le2\)

\(\Leftrightarrow\dfrac{3bc}{4a+3bc}+\dfrac{3ac}{4b+3ac}+\dfrac{3ab}{4c+3ab}\ge1\)

\(\Leftrightarrow\dfrac{bc}{4a+3bc}+\dfrac{ac}{4b+3ac}+\dfrac{ab}{4c+3ab}\ge\dfrac{1}{3}\)

Thật vậy, ta có:

\(VT=\dfrac{\left(bc\right)^2}{4abc+3\left(bc\right)^2}+\dfrac{\left(ca\right)^2}{4abc+3\left(ac\right)^2}+\dfrac{\left(ab\right)^2}{4abc+3\left(ab\right)^2}\)

\(VT\ge\dfrac{\left(ab+bc+ca\right)^2}{3\left(ab\right)^2+3\left(bc\right)^2+3\left(ca\right)^2+12abc}=\dfrac{\left(ab+bc+ca\right)^2}{3\left(ab\right)^2+3\left(bc\right)^2+3\left(ca\right)^2+6abc\left(a+b+c\right)}\)

\(VT\ge\dfrac{\left(ab+bc+ca\right)^2}{3\left(ab+bc+ca\right)^2}=\dfrac{1}{3}\) (đpcm)

Dấu "=" xảy ra khi \(a=b=c=...\)

Trừ mỗi vế cho 1, ta có:

\(\frac{b-16a+16c}{4a}=\frac{c-16b+16a}{4b}=\frac{a-16c+16b}{4c}=\frac{a+b+c}{4.\left(a+b+c\right)}=\frac{1}{4}\)(vì a,b,c > 0 nên a+b+c>0)

\(\Leftrightarrow\hept{\begin{cases}b+16c=17a\\c+16a=17b\\a+16b=17c\end{cases}}\Leftrightarrow a=b=c\)

tự thay vào

\(\frac{4a}{b}=\frac{4b}{c}=\frac{4c}{a}=\frac{4a+4b+4c}{b+c+a}=\frac{4\left(a+b+c\right)}{a+b+c}=4\)

=> 4b=4a =>b=a 

=> 4b=4c => b=c 

=> a=b=c