Ta có

\(\sqrt{2}\sqrt{4a+1}\le\frac{4a+3}{2}\)

\(\sqrt{2}\sqrt{4b+1}\le\frac{4b+3}{2}\)

\(\sqrt{2}\sqrt{4c+1}\le\frac{4c+3}{2}\)

\(\sqrt{2}\sqrt{4d+1}\le\frac{4d+3}{2}\)

Cộng vế theo vế ta được

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

\(\le8\)

<=> \(\sqrt{4a+1}+\sqrt{4b+1}+\sqrt{4c+1}\sqrt{4d+1}\le4\sqrt{2}\)

Đề bài thiếu, chắc chắn phải có thêm 1 dữ kiện khác

Ví dụ, bạn cho \(a=b=c=1000\) sẽ thấy BĐT sai

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

\(\sqrt{4a+1}+\sqrt{4b+1}+\sqrt{4c+1}\le\sqrt{1+1+1}.\sqrt{4\left(a+b+c\right)+3}=\sqrt{3.7}=\sqrt{21}\)

Dau '=' xay ra khi \(a=b=c=\frac{1}{3}\)

đặt \(S=\frac{a}{4b^2+1}+\frac{b}{4c^2+1}+\frac{c}{4a^2+1}\)

\(=\frac{a^3}{4a^2b^2+a^2}+\frac{b^3}{4b^2c^2+b^2}+\frac{c^3}{4a^2c^2+c^2}\ge\frac{\left(a\sqrt{a}+b\sqrt{b}+c\sqrt{c}\right)^2}{4a^2b^2+4b^2c^2+4c^2a^2+a^2+b^2+c^2}\)

xét hiệu:

1-4(a²b²+b²c²+c²a²)-a²-b²-c²

=2ab+2bc+2ca-4(a²b²+b²c²+c²a²)

=2ab(1-2ab)+2bc(1-2bc)+2ca(1-2ca)

ta có:

\(2ab\le\frac{\left(a+b\right)^2}{2}\le\frac{1}{2};2bc\le\frac{\left(b+c\right)^2}{2}\le\frac{1}{2};2ca\le\frac{\left(c+a\right)^2}{2}\le\frac{1}{2}\)

\(\Rightarrow2ab\left(1-2ab\right);2bc\left(1-2bc\right);2ca\left(1-2ca\right)\ge0\)

\(\Rightarrow1\ge4\left(a^2b^2+b^2c^2+c^2a^2\right)+a^2+b^2+c^2\)

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

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

=>đpcm

dấu"=" xảy ra khi 1 số=1;2 số còn lại =0

Ta có: \(4b\sqrt{c}-c\sqrt{a}=\sqrt{c}\left(4b-\sqrt{ac}\right)>0\)( do \(1< a,b,c< 2\))

Tương tự => Các MS dương

\(VT=\frac{ba}{4b\sqrt{ac}-ca}+\frac{cb}{4c\sqrt{ba}-ab}+\frac{ac}{4a\sqrt{bc}-bc}\)

Áp dụng BĐT cosi schawr ta có

\(VT\ge\frac{\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\right)^2}{4b\sqrt{ac}+4c\sqrt{ab}+4a\sqrt{bc}-ab-bc-ac}\)

Áp dụng cosi ta có \(2b\sqrt{ac}=2\sqrt{ab}.\sqrt{ac}\le ab+ac\);\(2c\sqrt{ab}\le ac+bc\);\(2a\sqrt{bc}\le ab+ac\)

=> \(VT\ge\frac{\left(\sqrt{ab}+\sqrt{ac}+\sqrt{bc}\right)^2}{ab+bc+ac+2\sqrt{ab}+2\sqrt{bc}+2\sqrt{ac}}=\frac{\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\right)^2}{\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\right)^2}=1\)(ĐPCM)

Dấu bằng xảy ra khi a=b=c

Đặt:

\(A=\sqrt{4a+1}+\sqrt{4b+1}+\sqrt{4c+1}\)

Áp dụng bất đẳng thức bunhiacopxki ta có:

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

Hay \(A\le\sqrt{21}\left(đpcm\right)\)

Rảnh quá ủng hộ cách khác nè =))

Áp dụng Cô-si có:

\(\sqrt{4a+1}\cdot\sqrt{\dfrac{7}{3}}\le\dfrac{4a+1+\dfrac{7}{3}}{2}=2a+\dfrac{5}{3}\)

Tương tự vs 2 bđt còn lại: \(\left\{{}\begin{matrix}\sqrt{4b+1}\cdot\sqrt{\dfrac{7}{3}}\le2b+\dfrac{5}{3}\\\sqrt{4c+1}\cdot\sqrt{\dfrac{7}{3}}\le2c+\dfrac{5}{3}\end{matrix}\right.\)

Cộng 2 vế của 3 bđt trên có:

\(\sqrt{\dfrac{7}{3}}\left(\sqrt{4a+1}+\sqrt{4b+1}+\sqrt{4c+1}\right)\le2\left(a+b+c\right)+5=7\)

\(\Leftrightarrow\sqrt{4a+1}+\sqrt{4b+1}+\sqrt{4c+1}\le\sqrt{21}\)

Hoàn tất chứng minh

Unruly KidNguyễn Huy TúNguyễn Huy Thắng

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

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

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

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

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

áp dụng bất đẳng thức: (a+b+c)^2<=3(a^2+b^2+c^2): 
[√(4a+1)+√(4b+1)+√(4c+1)]^2 
<= 3[4(a+b+c)+3]=21<25 
=>√(4a+1)+√(4b+1)+√(4c+1)<5

cosi : \(\sqrt{4a+1}\)\(\sqrt{1}\)<\(\frac{4a+1+1}{2}\)= 2a + 1. tương tự  \(\sqrt{4b+1}\)\(\sqrt{1}\)<\(\frac{4b+1+1}{2}\)= 2b + 1;  \(\sqrt{4c+1}\)\(\sqrt{1}\)<\(\frac{4c+1+1}{2}\)= 2c + 1. Nên VT < 2(a+b+c) +3 = 5. Dấu = xảy ra khi và chỉ khi a=b=c = 1/3