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

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.

Ta có: \(\left(a^4-a^3+2\right)-\left(a+1\right)=\left(a-1\right)^2\left(a^2+a+1\right)\ge0\)\(\Rightarrow a^4-a^3+2\ge a+1\Leftrightarrow a^4-a^3+ab+2\ge ab+a+1\)

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

Tương tự:\(\frac{1}{\sqrt{b^4-b^3+bc+2}}\le\frac{1}{\sqrt{bc+b+1}}\); \(\frac{1}{\sqrt{c^4-c^3+ca+2}}\le\frac{1}{\sqrt{ca+c+1}}\)

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

Đẳng thức xảy ra khi a = b = c = 1

câu bất này dễ mà Việt Anh

nhưng mình kém lắm

ko làm đc

giúp đi

Lời giải:
Vì $ab+bc+ac=1$ nên:

$a^2+1=a^2+ab+bc+ac=(a+b)(b+c)$

$b^2+1=b^2+ab+bc+ac=(b+a)(b+c)$

$c^2+1=c^2+ab+bc+ac=(c+a)(c+b)$

Do đó, áp dụng BĐT AM-GM:

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

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

Ta có đpcm

Dấu "=" xảy ra khi \(a=b=c=\frac{1}{\sqrt{3}}\)

Lời giải:
Vì $ab+bc+ac=1$ nên:

$a^2+1=a^2+ab+bc+ac=(a+b)(b+c)$

$b^2+1=b^2+ab+bc+ac=(b+a)(b+c)$

$c^2+1=c^2+ab+bc+ac=(c+a)(c+b)$

Do đó, áp dụng BĐT AM-GM:

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

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

Ta có đpcm

Dấu "=" xảy ra khi \(a=b=c=\frac{1}{\sqrt{3}}\)

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

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