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Lời giải:
Không mất tổng quát giả sử \(c=\min (a,b,c)\)
Khi đó, do \(ab+bc+ac=3\Rightarrow ab\geq 1\).
Với $ab\geq 1$ ta có bổ đề sau: \(\frac{1}{a^2+1}+\frac{1}{b^2+1}\geq \frac{2}{ab+1}\)
Để cm bổ đề trên rất đơn giản. Quy đồng và biến đổi tương đương thu được \((a-b)^2(ab-1)\geq 0\) (luôn đúng với mọi \(ab\geq 1\) )
Sử dụng bổ đề vào bài toán:
\(\Rightarrow \text{VT}\geq \frac{2}{ab+1}+\frac{1}{c^2+1}=\frac{2c^2+ab+3}{abc^2+ab+c^2+1}(*)\)
Giờ ta sẽ cm \(\frac{2c^2+ab+3}{abc^2+ab+c^2+1}\geq \frac{3}{2}(**)\)
\(\Leftrightarrow 2(2c^2+ab+3)\geq 3(abc^2+ab+c^2+1)\)
\(\Leftrightarrow c^2+3\geq 3abc^2+ab\)
\(\Leftrightarrow c^2+bc+ac\geq 3abc^2\)
\(\Leftrightarrow c+b+a\geq 3abc\).
BĐT trên đúng do theo AM-GM: \(3(a+b+c)=(ab+bc+ac)(a+b+c)\geq 9abc\Rightarrow a+b+c\geq 3abc\) )
Do đó $(*)$ được cm.
Từ \((*),(**)\Rightarrow \text{VT}\geq \frac{3}{2}\) (đpcm)
Dấu bằng xảy ra khi $a=b=c=1$
Đặt \(\left(a,b,c\right)\rightarrow\left(\dfrac{x}{y},\dfrac{y}{z},\dfrac{z}{x}\right)\)
BĐT cần c/m tương đương với
\(\sum\dfrac{yz}{xy+xz+2yz}\le\dfrac{3}{4}\)
\(\Leftrightarrow\sum\dfrac{xy+xz}{xy+xz+2yz}\ge\dfrac{3}{2}\)
Ta có \(\sum\dfrac{xy+xz}{xy+xz+2yz}\ge\dfrac{\left(2\sum xy\right)^2}{\sum\left(xy+xz+2yz\right)\left(xy+xz\right)}=\dfrac{4\left(\sum xy\right)^2}{2\sum x^2y^2+6\sum x^2yz}\)
Như vậy ta cần c/m \(\dfrac{4\left(\sum xy\right)^2}{2\sum x^2y^2+6\sum x^2yz}\ge\dfrac{3}{2}\)
\(\Leftrightarrow8\left(\sum xy\right)^2\ge6\sum x^2y^2+18\sum x^2yz\)
\(\Leftrightarrow8\left(\sum xy\right)^2\ge6\left(\sum xy\right)^2+6\sum x^2yz\)
\(\Leftrightarrow\left(\sum xy\right)^2\ge3\sum x^2yz\) (luôn đúng)
Ta có:
\(\dfrac{1}{ab+a+2}\le\dfrac{1}{4}\left(\dfrac{1}{ab+1}+\dfrac{1}{a+1}\right)=\dfrac{1}{4}\left(\dfrac{c}{1+c}+\dfrac{1}{a+1}\right)\)
Tương tự cho 2 BĐT còn lại rồi cộng theo vế:
\(VT\le\dfrac{1}{4}\left(\dfrac{a+1}{a+1}+\dfrac{b+1}{b+1}+\dfrac{c+1}{c+1}\right)=\dfrac{3}{4}\)
\(VT=\dfrac{a^3}{a^2+abc}+\dfrac{b^3}{b^2+abc}+\dfrac{c^3}{c^2+abc}\)
Xét \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=1\Leftrightarrow ab+bc+ac=abc\)
\(\Rightarrow VT=\dfrac{a^3}{a^2+ab+bc+ac}+\dfrac{b^3}{b^2+ab+bc+ac}+\dfrac{c^3}{c^2+ab+bc+ac}\)
\(\Leftrightarrow VT=\dfrac{a^3}{\left(a+b\right)\left(a+c\right)}+\dfrac{b^3}{\left(b+a\right)\left(b+c\right)}+\dfrac{c^3}{\left(c+b\right)\left(c+a\right)}\)
Áp dụng bđt Cauchy ta có :
\(\dfrac{a^3}{\left(a+b\right)\left(a+c\right)}+\dfrac{a+b}{8}+\dfrac{a+c}{8}\ge3\sqrt[3]{\dfrac{a^3}{64}}=\dfrac{3a}{4}\)
Thiết lập tương tự và thu lại ta có :
\(VT+\dfrac{a+b+c}{2}\ge\dfrac{3}{4}\left(a+b+c\right)\)
\(\Rightarrow VT\ge\dfrac{3}{4}\left(a+b+c\right)-\dfrac{1}{2}\left(a+b+c\right)=\dfrac{a+b+c}{4}\left(đpcm\right)\)
Dấu '' = '' xảy ra khi \(a=b=c=3\)
Có BĐT: \(a^2+b^2+c^2\ge ab+bc+ca\)
\(\Leftrightarrow\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\)
Ta có:
\(VT=\)\(\dfrac{1}{a^2+b^2+1}+\dfrac{1}{b^2+c^2+1}+\dfrac{1}{c^2+a^2+1}\)
\(=\dfrac{1+1+c^2}{\left(a^2+b^2+1\right)\left(1+1+c^2\right)}+\dfrac{1+1+a^2}{\left(b^2+c^2+1\right)\left(1+1+a^2\right)}+\dfrac{1+1+b^2}{\left(c^2+a^2+1\right)\left(1+1+b^2\right)}\)
Áp dụng BĐT Bunhiacopski cho mẫu số, ta có:
\(\left(a^2+b^2+c^2\right)\left(1+1+c^2\right)\ge\left(a+b+c\right)^2\)
\(\left(b^2+c^2+1\right)\left(1+1+a^2\right)\ge\left(b+c+a\right)^2\)
\(\left(c^2+a^2+1\right)\left(1+1+b^2\right)\ge\left(c+a+b\right)^2\)
\(\Rightarrow VT\le\dfrac{1+1+c^2}{\left(a+b+c\right)^2}+\dfrac{1+1+a^2}{\left(b+c+a\right)^2}+\dfrac{1+1+b^2}{\left(c+a+b\right)^2}=\dfrac{6+a^2+b^2+c^2}{\left(a+b+c\right)^2}\le\dfrac{6+ab+bc+ca}{3\left(ab+bc+ca\right)}=\dfrac{6+3}{3.3}=1\)
\("="\Leftrightarrow a=b=c=1\)
Ta có a,b,c dương⇒\(a+b+c+ab+bc+ca=6abc\Leftrightarrow\dfrac{1}{cb}+\dfrac{1}{ac}+\dfrac{1}{ab}+\dfrac{1}{c}+\dfrac{1}{a}+\dfrac{1}{b}=6\)(1)
Đặt x=\(\dfrac{1}{a}\),y=\(\dfrac{1}{b}\),z=\(\dfrac{1}{c}\)
Vậy (1)\(\Leftrightarrow xy+xz+yz+x+y+z=6\)
Áp dụng bđt cosi ta có
\(x^2+1\ge2x\)(2)
\(y^2+1\ge2y\)(3)
\(z^2+1\ge2z\)(4)
Cộng (2),(3),(4)\(\Leftrightarrow x^2+y^2+z^2+3\ge2x+2y+2z\)(5)
Ta lại có bất đẳng thức cosi:
\(x^2+y^2\ge2xy\)(6)
\(y^2+z^2\ge2yz\)(7)
\(x^2+z^2\ge2xz\)(8)
Cộng (6),(7),(8)\(\Leftrightarrow2\left(x^2+y^2+z^2\right)\ge2xy+2xz+2yz\left(9\right)\)
Cộng (8),(9)\(\Leftrightarrow3\left(x^2+y^2+z^2\right)+3\ge2\left(x+y+z+xy+xz+yz\right)\Leftrightarrow3\left(x^2+y^2+z^2\right)+3\ge2.6\Leftrightarrow3\left(x^2+y^2+z^2\right)\ge9\Leftrightarrow x^2+y^2+z^2\ge3\Leftrightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\ge3\Rightarrowđpcm\)
Đặt A = \(\dfrac{a-b}{1+c^2}+\dfrac{b-c}{1+a^2}+\dfrac{c-a}{1+b^2}=0\)
= \(\dfrac{a-b}{c^2+ab+bc+ca}+\dfrac{b-c}{a^2+ab+bc+ca}+\dfrac{c-a}{b^2+ab+bc+ca}\)
= \(\dfrac{a-b}{\left(c+a\right)\left(c+b\right)}+\dfrac{b-c}{\left(a+b\right)\left(c+a\right)}+\dfrac{c-a}{\left(a+b\right)\left(b+c\right)}\)
= \(\dfrac{\left(a-b\right)\left(a+b\right)+\left(b-c\right)\left(b+c\right)+\left(c+a\right)\left(c-a\right)}{\left(c+a\right)\left(b+c\right)\left(a+b\right)}\)
= \(\dfrac{a^2-b^2+b^2-c^2+c^2-a^2}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}=0\)
\(\dfrac{a-b}{1+c^2}+\dfrac{b-c}{1+a^2}+\dfrac{c-a}{1+b^2}\)
\(=\dfrac{a-b}{ab+bc+ca+c^2}+\dfrac{b-c}{ab+bc+ca+a^2}+\dfrac{c-a}{ab+bc+ca+b^2}\)
\(=\dfrac{a-b}{\left(c+a\right)\left(c+b\right)}+\dfrac{b-c}{\left(a+b\right)\left(a+c\right)}+\dfrac{c-a}{\left(b+a\right)\left(b+c\right)}\)
\(=\dfrac{\left(a-b\right)\left(a+b\right)+\left(b-c\right)\left(b+c\right)+\left(c-a\right)\left(c+a\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)
\(=\dfrac{a^2-b^2+b^2-c^2+c^2-a^2}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}=0\)