\(\Sigma\frac{b+1}{8-\sqrt{a}}\le\Sigma\frac{2\left(b+1\right)}{15-a}=\Sigma\frac{2\left(a+2b+c\right)}{4a+5b+5c}\)(AM-gm)

Đặt \(\left\{\begin{matrix}x=4a+5b+5c\\y=4b+5a+5c\\z=4c+5a+5b\end{matrix}\right.\)suy ra...

tiếp đi?

\(\frac{1}{\sqrt{a^4-a^3+ab+2}}+\frac{1}{\sqrt{b^4-b^3+bc+2}}+\frac{1}{\sqrt{c^4-c^3+ca+2}}\)\(\left(a,b,c>0\right)\).

Với \(a,b>0\), ta có:

\(\left(a-1\right)^2\left(a^2+a+1\right)\ge0\).

\(\Leftrightarrow\left(a^3-1\right)\left(a-1\right)\ge0\).

\(\Leftrightarrow a^4-a^3-a+1\ge0\).

\(\Leftrightarrow a^4-a^3+1\ge a\).

\(\Leftrightarrow a^4-a^3+ab+2\ge ab+a+1\).

\(\Leftrightarrow\sqrt{a^4-a^3+ab+2}\ge\sqrt{ab+a+1}\).

\(\Rightarrow\frac{1}{\sqrt{a^4-a^3+ab+2}}\le\frac{1}{\sqrt{ab+a+1}}\left(1\right)\).

Dấu bằng xảy ra \(\Leftrightarrow a-1=0\Leftrightarrow a=1\).

Chứng minh tương tự (với \(b,c>0\)), ta được:

\(\frac{1}{\sqrt{b^4-b^3+bc+2}}\le\frac{1}{\sqrt{bc+b+1}}\left(2\right)\).

Dấu bằng xảy ra \(\Leftrightarrow b=1\).

Chứng minh tương tự (với \(a,c>0\)), ta được:

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

Dấu bằng xảy ra \(\Leftrightarrow c=1\).

Từ \(\left(1\right),\left(2\right),\left(3\right)\), ta được:

\(\frac{1}{\sqrt{a^4-a^3+ab+2}}+\frac{1}{\sqrt{b^4-b^3+bc+2}}+\frac{1}{\sqrt{c^4-c^3+ca+2}}\)\(\le\frac{1}{\sqrt{ab+a+1}}+\frac{1}{\sqrt{bc+b+1}}+\frac{1}{\sqrt{ca+c+1}}\left(4\right)\).

Áp dụng bất đẳng thức Bu-nhi-a-cốp-xki cho 3 số, ta được:

\(\left(1.\frac{1}{\sqrt{ab+a+1}}+1.\frac{1}{\sqrt{bc+b+1}}+1.\frac{1}{\sqrt{ca+c+1}}\right)^2\)\(\le\)\(\left(1^2+1^2+1^2\right)\)\(\left[\frac{1}{\left(\sqrt{ab+a+1}\right)^2}+\frac{1}{\left(\sqrt{bc+b+1}\right)^2}+\frac{1}{\left(\sqrt{ca+c+1}\right)^2}\right]\).

\(\Leftrightarrow\left(\frac{1}{\sqrt{ab+a+1}}+\frac{1}{\sqrt{bc+b+1}}+\frac{1}{\sqrt{ca+c+1}}\right)^2\)\(\le3\left(\frac{1}{ab+b+1}+\frac{1}{bc+b+1}+\frac{1}{ca+c+1}\right)\).

Ta có:

\(\frac{1}{ab+a+1}+\frac{1}{bc+b+1}+\frac{1}{ca+c+1}\)

\(=\frac{c}{abc+ac+c}+\frac{abc}{bc+b+abc}+\frac{1}{ca+c+1}\)(vì \(abc=1\)).

\(=\frac{c}{1+ac+c}+\frac{abc}{b\left(c+1+ac\right)}+\frac{1}{ca+c+1}\)(vì \(abc=1\)).

\(=\frac{c}{1+ac+c}+\frac{ac}{1+ac+c}+\frac{1}{1+ac+c}=1\).

Do đó:

\(\left(\frac{1}{\sqrt{ab+a+1}}+\frac{1}{\sqrt{bc+b+1}}+\frac{1}{\sqrt{ca+c+1}}\right)^2\le3.1=3\).

\(\Leftrightarrow\frac{1}{\sqrt{ab+a+1}}+\frac{1}{\sqrt{bc+b+1}}+\frac{1}{\sqrt{ca+c+1}}\le\sqrt{3}\left(5\right)\).

Từ \(\left(4\right)\)và \(\left(5\right)\), ta được:

\(\frac{1}{\sqrt{a^4-a^3+ab+2}}+\frac{1}{\sqrt{b^4-b^3+bc+2}}+\frac{1}{\sqrt{c^4-c^3+ca+2}}\le\)\(\sqrt{3}\)(điều phải chứng minh).
Dấu bằng xảy ra \(\Leftrightarrow a=b=c=1\).

Vậy \(\frac{1}{\sqrt{a^4-a^3+ab+2}}+\frac{1}{\sqrt{b^4-b^3+bc+2}}+\frac{1}{\sqrt{c^4-c^3+ca+2}}\)\(\le\sqrt{3}\)với \(a,b,c>0\)và \(abc=1\).

\(+2\)nhé, không phải \(-2\)đâu.

Áp dụng giả thiết và một đánh giá quen thuộc, ta được: \(16\left(a+b+c\right)\ge\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{ab+bc+ca}{abc}=\frac{\left(ab+bc+ca\right)^2}{abc\left(ab+bc+ca\right)}\ge\frac{3\left(a+b+c\right)}{ab+bc+ca}\)hay \(\frac{1}{6\left(ab+bc+ca\right)}\le\frac{8}{9}\)

Đến đây, ta cần chứng minh \(\frac{1}{\left(a+b+\sqrt{2\left(a+c\right)}\right)^3}+\frac{1}{\left(b+c+\sqrt{2\left(b+a\right)}\right)^3}+\frac{1}{\left(c+a+\sqrt{2\left(c+b\right)}\right)^3}\le\frac{1}{6\left(ab+bc+ca\right)}\)

 Áp dụng bất đẳng thức Cauchy cho ba số dương ta có \(a+b+\sqrt{2\left(a+c\right)}=a+b+\sqrt{\frac{a+c}{2}}+\sqrt{\frac{a+c}{2}}\ge3\sqrt[3]{\frac{\left(a+b\right)\left(a+c\right)}{2}}\)hay \(\left(a+b+\sqrt{2\left(a+c\right)}\right)^3\ge\frac{27\left(a+b\right)\left(a+c\right)}{2}\Leftrightarrow\frac{1}{\left(a+b+2\sqrt{a+c}\right)^3}\le\frac{2}{27\left(a+b\right)\left(a+c\right)}\)

Hoàn toàn tương tự ta có \(\frac{1}{\left(b+c+2\sqrt{b+a}\right)^3}\le\frac{2}{27\left(b+c\right)\left(b+a\right)}\); \(\frac{1}{\left(c+a+2\sqrt{c+b}\right)^3}\le\frac{2}{27\left(c+a\right)\left(c+b\right)}\)

Cộng theo vế các bất đẳng thức trên ta được \(\frac{1}{\left(a+b+\sqrt{2\left(a+c\right)}\right)^3}+\frac{1}{\left(b+c+\sqrt{2\left(b+a\right)}\right)^3}+\frac{1}{\left(c+a+\sqrt{2\left(c+b\right)}\right)^3}\le\frac{4\left(a+b+c\right)}{27\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)Phép chứng minh sẽ hoàn tất nếu ta chỉ ra được \(\frac{4\left(a+b+c\right)}{27\left(a+b\right)\left(b+c\right)\left(c+a\right)}\le\frac{1}{6\left(ab+bc+ca\right)}\)\(\Leftrightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge\frac{8}{9}\left(ab+bc+ca\right)\left(a+b+c\right)\)

Đây là một đánh giá đúng, thật vậy: đặt a + b + c = p; ab + bc + ca = q; abc = r thì bất đẳng thức trên trở thành \(pq-r\ge\frac{8}{9}pq\Leftrightarrow\frac{1}{9}pq\ge r\)*đúng vì \(a+b+c\ge3\sqrt[3]{abc}\); \(ab+bc+ca\ge3\sqrt[3]{\left(abc\right)^2}\))

Vậy bất đẳng thức được chứng minh

Đẳng thức xảy ra khi \(a=b=c=\frac{1}{4}\)

Ta có 

\(\sqrt{\frac{ab}{c+ab}}=\sqrt{\frac{ab}{c\left(a+b+c\right)+ab}}\)\(=\sqrt{\frac{ab}{\left(c+a\right)\left(c+b\right)}}\)\(=\sqrt{\frac{a}{c+a}}.\sqrt{\frac{b}{c+b}}\)\(\le\frac{1}{2}\left(\frac{a}{c+a}+\frac{b}{c+b}\right)\)

Tương tự, ta có

\(\sqrt{\frac{bc}{a+bc}}\le\frac{1}{2}\left(\frac{b}{a+b}+\frac{c}{a+c}\right)\)

\(\sqrt{\frac{ca}{b+ca}\le\frac{1}{2}\left(\frac{c}{c+b}+\frac{a}{b+a}\right)}\)

Cộng vế theo vế của 3 bđt ta được đpcm

Từ giả thiết suy ra \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=3\)  (*) (Vì a,b,c > 0)

Áp dụng BĐT Cauchy ta có:

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

Đánh giá tương tự: \(\frac{1}{\sqrt{b^3+c}}\le\frac{1}{4\sqrt{2}}\left(\frac{3}{b}+\frac{1}{c}\right);\frac{1}{\sqrt{c^3+a}}\le\frac{1}{4\sqrt{2}}\left(\frac{3}{c}+\frac{1}{a}\right)\)

Từ đó, kết hợp với (*) suy ra:

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

Dấu "=" xảy ra khi và chỉ khi \(a=b=c=1.\)

kết bạn với mình không

Vì a,b,c là số thực dương nên \(\sqrt{a^2}=a;\sqrt{b^2}=b;\sqrt{c^2}\)=c. Vậy ta có

\(\frac{a}{a+1}+\frac{b}{b+1}+\frac{c}{c+1}\)=\(\frac{a}{a+1}-1+\frac{b}{b+1}-1\)+\(\frac{c}{c+1}-1+3\) 

=3-(  \(\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\)) =A

ta có bdt  \(9\le\left(a+1+b+1+c+1\right)\left(\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\right)\)(dễ dàng chứng mình bằng bdt cosi).

=>\(\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\ge\)\(\frac{9}{3+\sqrt{3}}\)=> A\(\le3-\frac{9}{3+\sqrt{3}}=\frac{3\sqrt{3}}{3+\sqrt{3}}=\frac{3}{\sqrt{3}+1}\)

dấu = khi a=b=c=\(\frac{\sqrt{3}}{3}\)

\(\frac{a}{\sqrt{1+a^2}}+\frac{b}{\sqrt{1+b^2}}+\frac{c}{\sqrt{1+c^2}}\)

\(=\frac{a}{\sqrt{\left(ab+bc+ca\right)+a^2}}+\frac{b}{\sqrt{\left(ab+bc+ca\right)+b^2}}+\frac{c}{\sqrt{\left(ab+bc+ca\right)+c^2}}\)

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

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

Ta có:\(a^5+ab+b^2\ge3a^2b\)

Tương tự ta có:

\(VT\le\frac{1}{\sqrt{3ab\left(a+2c\right)}}+\frac{1}{\sqrt{3bc\left(b+2a\right)}}+\frac{1}{\sqrt{3ca\left(c+2b\right)}}\)

\(=\frac{1}{\sqrt{3}}\left(\sqrt{\frac{c}{c+2a}}+\sqrt{\frac{a}{b+2a}}+\sqrt{\frac{b}{2b+c}}\right)\)

Ta cũng có:\(a+2c=a+c+c\ge\frac{1}{3}\left(\sqrt{a}+2\sqrt{c}\right)^2\)

\(\Rightarrow VT\le\frac{\sqrt{c}}{\sqrt{a}+2\sqrt{c}}+\frac{\sqrt{a}}{\sqrt{b}+2\sqrt{a}}+\frac{\sqrt{b}}{\sqrt{c}+2\sqrt{b}}\)

Đặt \(x=\frac{\sqrt{a}}{\sqrt{c}};y=\frac{\sqrt{b}}{\sqrt{a}};z=\frac{\sqrt{c}}{\sqrt{b}};xyz=1\)

\(\Rightarrow VT\le\frac{1}{x+2}+\frac{1}{y+2}+\frac{1}{z+2}\)

Giả sử \(xy\le1\) thì \(z\ge1\)

Ta có: \(\frac{1}{x+2}+\frac{1}{y+2}+\frac{1}{z+2}=\frac{1}{2}\left(\frac{1}{\frac{x}{2}+1}+\frac{1}{\frac{y}{2}+1}\right)+\frac{1}{z+2}\)

\(\le\frac{1}{1\frac{\sqrt{xy}}{2}}+\frac{1}{z+2}\le1\)(Đpcm)

Dấu = khi \(a=b=c=1\)

sao chứng minh đc \(a^5+ab+b^2\ge3a^2b\)vậy bạn