Đặt bđt là (*)

Để (*) đúng với mọi số thực dương a,b,c thỏa mãn :

\(a+b+c\ge\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)thì \(a=b=c=1\) cũng thỏa mãn (*)

\(\Rightarrow4\le\sqrt[n]{\left(n+2\right)^2}\)

Mặt khác: \(\sqrt[n]{\left(n+2\right)\left(n+2\right).1...1}\le\frac{2n+4+\left(n-2\right)}{n}=3+\frac{2}{n}\)

Hay \(n\le2\)

Với n=2 . Thay vào (*) : ta cần CM BĐT 

\(\frac{1}{\left(2a+b+c\right)^2}+\frac{1}{\left(2b+c+a\right)^2}+\frac{1}{\left(2c+a+b\right)^2}\le\frac{3}{16}\)

Với mọi số thực dương a,b,c thỏa mãn: \(a+b+c\ge\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)

Vì: \(\frac{1}{\left(2a+b+c\right)^2}\le\frac{1}{4\left(a+b\right)\left(a+c\right)}\)

Tương tự ta có:

\(\frac{1}{\left(2b+a+c\right)^2}\le\frac{1}{4\left(a+b\right)\left(a+c\right)};\frac{1}{\left(2c+a+b\right)^2}\le\frac{1}{4\left(a+c\right)\left(c+b\right)}\)

Ta cần CM: 

\(\frac{a+b+c}{2\left(a+b\right)\left(b+c\right)\left(c+a\right)}\le\frac{3}{16}\Leftrightarrow16\left(a+b+c\right)\le6\left(a+b\right)\left(b+c\right)\left(c+a\right)\)

Ta có BĐT: \(9\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge8\left(a+b+c\right)\left(ab+bc+ca\right)\)

Và: \(3\left(ab+cb+ac\right)\le3abc\left(a+b+c\right)\le\left(ab+cb+ca\right)^2\Rightarrow ab+bc+ca\ge3\)

=> đpcm

Dấu '=' xảy ra khi a=b=c

=> số nguyên dương lớn nhất : n=2( thỏa mãn)

Xét biểu thức \(\frac{1}{a+2}+\frac{1}{b+2}+\frac{1}{c+2}\)

\(=\frac{\left(a+2\right)\left(b+2\right)+\left(b+2\right)\left(c+2\right)+\left(c+2\right)\left(a+2\right)}{\left(a+2\right)\left(b+2\right)\left(c+2\right)}\)

\(=\frac{\left(ab+bc+ca\right)+4\left(a+b+c\right)+12}{abc+2\left(ab+bc+ca\right)+4\left(a+b+c\right)+8}\)

\(=\frac{\left(ab+bc+ca\right)+4\left(a+b+c\right)+12}{\left(abc+ab+bc+ca\right)+\left(ab+bc+ca\right)+4\left(a+b+c\right)+8}\)

\(=\frac{\left(ab+bc+ca\right)+4\left(a+b+c\right)+12}{4+\left(ab+bc+ca\right)+4\left(a+b+c\right)+8}\)(Do \(ab+bc+ca+abc=4\)theo giả thiết)

\(=\frac{\left(ab+bc+ca\right)+4\left(a+b+c\right)+12}{\left(ab+bc+ca\right)+4\left(a+b+c\right)+12}=1\)(***)

Với x,y dương ta có 2 bất đẳng thức phụ sau:

\(2\left(x^2+y^2\right)\ge\left(x+y\right)^2\)(*)

\(\frac{1}{x+y}\le\frac{1}{4}\left(\frac{1}{x}+\frac{1}{y}\right)\)(**)

Áp dụng (*) và (**), ta có:

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

\(\le\frac{1}{4}\left(\frac{1}{a+2}+\frac{1}{b+2}\right)\)(1)

Tương tự ta có: \(\frac{1}{\sqrt{2\left(b^2+c^2\right)}+4}\le\frac{1}{4}\left(\frac{1}{b+2}+\frac{1}{c+2}\right)\)(2)

\(\frac{1}{\sqrt{2\left(c^2+a^2\right)}+4}\le\frac{1}{4}\left(\frac{1}{c+2}+\frac{1}{a+2}\right)\)(3)

Cộng từng vế của các bất đẳng thức (1), (2), (3), ta được:

\(P\le\frac{1}{2}\left(\frac{1}{a+2}+\frac{1}{b+2}+\frac{1}{c+2}\right)=\frac{1}{2}\)(theo (***))

Đẳng thức xảy ra khi \(a=b=c\)

Bạn bổ sung cho mình dòng cuối là a = b = c = 1 nhé!

Hì , giải đc rùi nha.

Vì \(x,y\in R\)

\(\Rightarrow\left(x+2\right).\left(y+2\right)=\frac{25}{4}\)

Min \(P=\sqrt{1+x^4}+\sqrt{1+y^4}\)

- Dự đoán \(x=y=\frac{1}{2}\)

- Sử dụng BĐT : \(\frac{x^2}{a}+\frac{y^2}{b}\ge\frac{\left(x+y\right)^2}{a+b}\)    ( Với a,b > 0 )

=>  \(1+x^4=16.\frac{1}{16}+a^4=16.\left(\frac{1}{4}\right)^2+a^2\ge\frac{[16.\frac{1}{4}+a^2]^2}{17}\)

\(=\frac{(a^2+4)^2}{17}\)

=> \(1+y^4\ge\frac{\left(y^2+4\right)^2}{17}\)

=> \(P\ge\frac{x^2+y^2+8}{\sqrt{17}}\)

\(\Leftrightarrow P\sqrt{17}=\frac{1}{5}\left(x^2+y^2\right)+\frac{4}{5}\left(x^2+\frac{1}{4}+y^2+\frac{1}{4}\right)+8-\frac{2}{5}\)

\(\ge\frac{2xy}{5}+\frac{4}{5}\left(x+y\right)+8-\frac{2}{5}=\frac{2}{5}[xy+2\left(x+y\right)]+8-\frac{2}{5}\)

Theo giả thiết \(\left(x+2\right)\left(y+2\right)=\frac{25}{4}\)

\(\Leftrightarrow xy+2\left(x+y\right)=\frac{9}{4}\)

\(\Rightarrow P\sqrt{17}\ge\frac{2}{5}.\frac{9}{4}+8-\frac{2}{5}=\frac{17}{2}\)

\(\Leftrightarrow P\ge\frac{\sqrt{17}}{2}\)

Điểm rơi \(x=y=\frac{1}{2}\)

Ta có: 

\(\left(b-\dfrac{1}{2}\right)^2\ge0\) <=> \(b^2-b+\dfrac{1}{4}\ge0\) <=>\(b-\dfrac{1}{4}\le b^2\)

Mà : 

a<1 => \(log_a\left(b-\dfrac{1}{4}\right)\ge log_ab^2=2log_ab\)

P=\(log_a\left(b-\dfrac{1}{4}\right)-\dfrac{1}{2}log_{\dfrac{a}{b}}b=log_a\left(b-\dfrac{1}{4}\right)-\dfrac{1}{2}.\dfrac{log_ab}{1-log_ab}\ge2log_ab-\dfrac{1}{2}.\dfrac{log_ab}{1-log_ab}\)

Đặt t=log_ab

Do b<a<1 => t=log_ab >1

Khi đó \(P\ge2t+\dfrac{t}{2t-2}=f\left(t\right)\). Khảo sát f(t) trên (1;+\(\infty\)) ta đc

P\(\ge\)f(t) \(\ge\) f\(\left(\dfrac{3}{2}\right)\) = \(\dfrac{9}{2}\)

TK: Tìm Min (x^4 + 1) (y^4 + 1) với x + y = căn10 ; x , y > 0 - Thanh Truc

\(Q=\sum\dfrac{\left(a+b\right)^2}{\sqrt{2\left(b+c\right)^2+bc}}\ge\sum\dfrac{\left(a+b\right)^2}{\sqrt{2\left(b+c\right)^2+\dfrac{1}{4}\left(b+c\right)^2}}=\dfrac{2}{3}\sum\dfrac{\left(a+b\right)^2}{b+c}\)

\(Q\ge\dfrac{2}{3}.\dfrac{\left(a+b+b+c+c+a\right)^2}{a+b+b+c+c+a}=\dfrac{4}{3}\left(a+b+c\right)=\dfrac{4}{3}\)