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Đẳng thức quen thuộc: \(a^2+ab+bc+ca=\left(a+b\right)\left(a+c\right)\) và tương tự cho các mẫu số còn lại
Ta có:
\(\sum\dfrac{1}{a^2+1}=\sum\dfrac{1}{\left(a+b\right)\left(a+c\right)}=\dfrac{2\left(a+b+c\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}=\dfrac{2\left(ab+bc+ca\right)\left(a+b+c\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)
Mặt khác:
\(2\left(ab+bc+ca\right)\left(a+b+c\right)=\left[a\left(b+c\right)+b\left(c+a\right)+c\left(a+b\right)\right]\left(a+b+c\right)\)
\(\ge\left(a\sqrt{b+c}+b\sqrt{c+a}+c\sqrt{a+b}\right)^2\) (Bunhiacopxki)
\(\Rightarrow\sum\dfrac{1}{a^2+1}\ge\dfrac{\left(a\sqrt{b+c}+b\sqrt{c+a}+c\sqrt{a+b}\right)^2}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)
\(=\left(\dfrac{a}{\sqrt{\left(a+b\right)\left(b+c\right)}}+\dfrac{b}{\sqrt{\left(a+b\right)\left(b+c\right)}}+\dfrac{c}{\sqrt{\left(a+c\right)\left(b+c\right)}}\right)^2\)
\(=\left(\dfrac{a}{\sqrt{a^2+1}}+\dfrac{b}{\sqrt{b^2+1}}+\dfrac{c}{\sqrt{c^2+1}}\right)^2\)
Do đó ta chỉ cần chứng minh:
\(\dfrac{a}{\sqrt{a^2+1}}+\dfrac{b}{\sqrt{b^2+1}}+\dfrac{c}{\sqrt{c^2+1}}\le\dfrac{3}{2}\)
\(\Leftrightarrow\dfrac{a}{\sqrt{\left(a+b\right)\left(a+c\right)}}+\dfrac{b}{\sqrt{\left(a+b\right)\left(b+c\right)}}+\dfrac{c}{\sqrt{\left(a+c\right)\left(b+c\right)}}\le\dfrac{3}{2}\)
Đúng theo AM-GM:
\(\sum\dfrac{a}{\sqrt{\left(a+b\right)\left(a+c\right)}}\le\dfrac{1}{2}\sum\left(\dfrac{a}{a+b}+\dfrac{a}{a+c}\right)=\dfrac{3}{2}\)
Dấu "=" xảy ra khi \(a=b=c=\dfrac{1}{\sqrt{3}}\)
\(ab+1\le b\Rightarrow a+\dfrac{1}{b}\le1\)
Đặt \(\left(a;\dfrac{1}{b}\right)=\left(x;y\right)\Rightarrow x+y\le1\)
\(P=x+\dfrac{1}{x^2}+y+\dfrac{1}{y^2}=\left(\dfrac{x}{2}+\dfrac{x}{2}+\dfrac{1}{16x^2}\right)+\left(\dfrac{y}{2}+\dfrac{y}{2}+\dfrac{1}{16y^2}\right)+\dfrac{15}{16}\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}\right)\)
\(P\ge3\sqrt[3]{\dfrac{x^2}{64x^2}}+3\sqrt[3]{\dfrac{y^2}{64y^2}}+\dfrac{15}{32}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)^2\)
\(P\ge\dfrac{3}{2}+\dfrac{15}{32}\left(\dfrac{4}{x+y}\right)^2\ge\dfrac{3}{2}+\dfrac{15}{32}.\left(\dfrac{4}{1}\right)^2=9\)
Dấu "=" xảy ra khi \(x=y=\dfrac{1}{2}\) hay \(\left(a;b\right)=\left(\dfrac{1}{2};2\right)\)
1.
Nhân 2 vế của BĐT với \(\left(a+b+c\right)\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
\(3(a^2+b^2+c^2)(a+b)(b+c)(c+a)\ge(a+b+c)\left(Σ_{cyc}(a^2+b^2)(c+a)(c+b)\right)\)
\(\LeftrightarrowΣ_{perms}a^2b\left(a-b\right)^2\ge0\) *đúng*
Lời giải:
BĐT cần CM tương đương với:
\(\left[\frac{(a+b)(1-ab)}{(a^2+1)(b^2+1)}\right]^2\leq \frac{1}{4}\)
Đặt $a+b=x; ab=y$ thì BĐT \(\Leftrightarrow \left(\frac{x(1-y)}{y^2+x^2-2y+1}\right)^2=\left(\frac{x(y-1)}{x^2+(y-1)^2}\right)^2\leq \frac{1}{4}\)
Điều này luôn đúng vì theo BĐT AM-GM:
\([x^2+(y-1)^2]^2=x^4+(y-1)^4+2x^2(y-1)^2\geq 2x^2(y-1)^2+2x^2(y-1)^2=[2x(y-1)]^2\)
\(\Rightarrow \frac{[x(y-1)]^2}{[x^2+(y-1)^2]^2}\leq \frac{[x(y-1)]^2}{[2x(y-1)]^2}=\frac{1}{4}\)