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\(\dfrac{a^2b^2}{2a^2+b^2+3a^2b^2}=\dfrac{a^2b^2}{\left(a^2+b^2\right)+\left(a^2+a^2b^2\right)+2a^2b^2}\le\dfrac{a^2b^2}{2ab+2a^2b+2a^2b^2}=\dfrac{ab}{2\left(1+a+ab\right)}\)
Tương tự và cộng lại;
\(P\le\dfrac{1}{2}\left(\dfrac{ab}{1+a+ab}+\dfrac{bc}{1+b+bc}+\dfrac{ca}{1+c+ca}\right)\)
\(P\le\dfrac{1}{2}\left(\dfrac{ab}{1+a+ab}+\dfrac{abc}{a+ab+abc}+\dfrac{ab.ca}{ab+abc+ab.ca}\right)\)
\(P\le\dfrac{1}{2}\left(\dfrac{ab}{1+a+ab}+\dfrac{1}{a+ab+1}+\dfrac{a}{ab+1+a}\right)=\dfrac{1}{2}\)
Dấu "=" xảy ra khi \(a=b=c=1\)
\(\dfrac{\sqrt{b^2+a^2+a^2}}{ab}\ge\dfrac{\sqrt{\dfrac{1}{3}\left(b+a+a\right)^2}}{ab}=\dfrac{1}{\sqrt{3}}\left(\dfrac{1}{a}+\dfrac{2}{b}\right)\)
Tương tự: \(\dfrac{\sqrt{c^2+2b^2}}{bc}\ge\dfrac{1}{\sqrt{3}}\left(\dfrac{1}{b}+\dfrac{2}{c}\right)\) ; \(\dfrac{\sqrt{a^2+2c^2}}{ac}\ge\dfrac{1}{\sqrt{3}}\left(\dfrac{1}{c}+\dfrac{2}{a}\right)\)
Cộng vế với vế:
\(VT\ge\dfrac{1}{\sqrt{3}}\left(\dfrac{3}{a}+\dfrac{3}{b}+\dfrac{3}{c}\right)=\sqrt{3}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=1980\sqrt{3}\)
Dấu "=" xảy ra khi \(a=b=c=\dfrac{3}{1980}\)
Ta có: \(P\le\dfrac{a}{2a+2b+2}+\dfrac{b}{2b+2c+2}+\dfrac{c}{2c+2a+2}\)
Nên ta chỉ cần chứng minh:
\(\dfrac{a}{a+b+1}+\dfrac{b}{b+c+1}+\dfrac{c}{c+a+1}\le1\)
\(\Rightarrow\dfrac{a}{a+b+1}-1+\dfrac{b}{b+c+1}-1+\dfrac{c}{c+a+1}-1\le-2\)
\(\Leftrightarrow\dfrac{b+1}{a+b+1}+\dfrac{c+1}{b+c+1}+\dfrac{a+1}{c+a+1}\ge2\)
Thật vậy, ta có:
\(VT=\dfrac{\left(a+1\right)^2}{\left(a+1\right)\left(a+c+1\right)}+\dfrac{\left(b+1\right)^2}{\left(b+1\right)\left(a+b+1\right)}+\dfrac{\left(c+1\right)^2}{\left(c+1\right)\left(b+c+1\right)}\)
\(VT\ge\dfrac{\left(a+b+c+3\right)^2}{ab+bc+ca+3\left(a+b+c\right)+6}=\dfrac{2\left(ab+bc+ca\right)+6\left(a+b+c\right)+12}{ab+bc+ca+3\left(a+b+c\right)+6}=2\)
Dấu "=" xảy ra khi \(a=b=c=1\)
\(P=\dfrac{a^2}{ab+\dfrac{1}{b}}+\dfrac{b^2}{bc+\dfrac{1}{c}}+\dfrac{c^2}{ca+\dfrac{1}{a}}\ge\dfrac{\left(a+b+c\right)^2}{ab+bc+ca+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}}\)
\(P\ge\dfrac{3\left(ab+bc+ca\right)}{ab+bc+ca+\dfrac{ab+bc+ca}{abc}}=\dfrac{3}{1+\dfrac{1}{abc}}=\dfrac{3abc}{1+abc}\) (đpcm)
Dấu "=" xảy ra khi \(a=b=c\)
Với a, b, c > 0 có:
\(P=\dfrac{a}{b+2c}+\dfrac{b}{c+2a}+\dfrac{c}{a+2b}\\ =\dfrac{a^2}{a\left(b+2c\right)}+\dfrac{b^2}{b\left(c+2a\right)}+\dfrac{c^2}{c\left(a+2b\right)}\)
\(\Rightarrow P\ge\dfrac{\left(a+b+c\right)^2}{\left(1+\alpha\right)\left(ab+bc+ca\right)}\ge\dfrac{3\left(ab+bc+ca\right)}{\left(1+\alpha\right)\left(ab+bc+ca\right)}\)
chọn \(\alpha=\dfrac{1}{abc}\Rightarrow dpcm\)
Áp dụng bất đẳng thức Cauchy-Schwarz ta có:
\(\dfrac{1}{2a^2+b^2}=\dfrac{1}{a^2+a^2+b^2}\le\dfrac{1}{9}\left(\dfrac{1}{a^2}+\dfrac{1}{a^2}+\dfrac{1}{b^2}\right)\)
\(\left\{{}\begin{matrix}\dfrac{1}{2b^2+c^2}\le\dfrac{1}{9}\left(\dfrac{1}{b^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)\\\dfrac{1}{2c^2+a^2}\le\dfrac{1}{9}\left(\dfrac{1}{c^2}+\dfrac{1}{c^2}+\dfrac{1}{a^2}\right)\end{matrix}\right.\)
Cộng theo vế:
\(L\le\dfrac{1}{9}\left(\dfrac{3}{a^2}+\dfrac{3}{b^2}+\dfrac{3}{c^2}\right)=\dfrac{1}{9}\left[3\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)\right]=\dfrac{1}{9}\)
Lời giải:
Áp dụng BĐT Cauchy ta có:
\(a^2+b^2\geq 2ab\)
\(b^2+1\geq 2b\)
Suy ra \(a^2+2b^2+3\geq 2(ab+b+1)\) \(\Rightarrow \frac{1}{a^2+2b^2+3}\leq \frac{1}{2(ab+b+1)}\)
Thực hiện toàn toàn tương tự với các phân thức còn lại và cộng theo vế:
\(\text{VT}\leq \frac{1}{2}\underbrace{\left(\frac{1}{ab+b+1}+\frac{1}{bc+c+1}+\frac{1}{ca+a+1}\right)}_{M}(1)\)
Lại có: \(M=\frac{1}{ab+b+1}+\frac{1}{bc+c+1}+\frac{1}{ca+a+1}=\frac{ac}{ab.ac+b.ac+ac}+\frac{a}{bc.a+c.a+a}+\frac{1}{ca+a+1}\)
\(=\frac{ac}{a+1+ac}+\frac{a}{1+ac+a}+\frac{1}{ac+a+1}=\frac{ac+a+1}{ac+a+1}=1(2)\)
Từ \((1); (2)\Rightarrow \text{VT}\leq \frac{1}{2}\) (đpcm)
Dấu bằng xảy ra khi \(a=b=c=1\)
Tham khảo nhá