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\(a^2+b^2-ab\ge\dfrac{1}{2}\left(a+b\right)^2-\dfrac{1}{4}\left(a+b\right)^2=\dfrac{1}{4}\left(a+b\right)^2\)
\(\Rightarrow\dfrac{1}{\sqrt{a^2-ab+b^2}}\le\dfrac{1}{\sqrt{\dfrac{1}{4}\left(a+b\right)^2}}=\dfrac{2}{a+b}\le\dfrac{1}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\)
Tương tự:
\(\dfrac{1}{\sqrt{b^2-bc+c^2}}\le\dfrac{1}{2}\left(\dfrac{1}{b}+\dfrac{1}{c}\right)\) ; \(\dfrac{1}{\sqrt{c^2-ca+a^2}}\le\dfrac{1}{2}\left(\dfrac{1}{c}+\dfrac{1}{a}\right)\)
Cộng vế:
\(P\le\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=3\)
Dấu "=" xảy ra khi \(a=b=c=1\)
Xét biểu thức \(\frac{1}{a+2}+\frac{1}{b+2}+\frac{1}{c+2}\)
\(=\frac{\left(a+2\right)\left(b+2\right)+\left(b+2\right)\left(c+2\right)+\left(c+2\right)\left(a+2\right)}{\left(a+2\right)\left(b+2\right)\left(c+2\right)}\)
\(=\frac{\left(ab+bc+ca\right)+4\left(a+b+c\right)+12}{abc+2\left(ab+bc+ca\right)+4\left(a+b+c\right)+8}\)
\(=\frac{\left(ab+bc+ca\right)+4\left(a+b+c\right)+12}{\left(abc+ab+bc+ca\right)+\left(ab+bc+ca\right)+4\left(a+b+c\right)+8}\)
\(=\frac{\left(ab+bc+ca\right)+4\left(a+b+c\right)+12}{4+\left(ab+bc+ca\right)+4\left(a+b+c\right)+8}\)(Do \(ab+bc+ca+abc=4\)theo giả thiết)
\(=\frac{\left(ab+bc+ca\right)+4\left(a+b+c\right)+12}{\left(ab+bc+ca\right)+4\left(a+b+c\right)+12}=1\)(***)
Với x,y dương ta có 2 bất đẳng thức phụ sau:
\(2\left(x^2+y^2\right)\ge\left(x+y\right)^2\)(*)
\(\frac{1}{x+y}\le\frac{1}{4}\left(\frac{1}{x}+\frac{1}{y}\right)\)(**)
Áp dụng (*) và (**), ta có:
\(\frac{1}{\sqrt{2\left(a^2+b^2\right)}+4}\le\frac{1}{a+b+4}=\frac{1}{\left(a+2\right)+\left(b+2\right)}\)
\(\le\frac{1}{4}\left(\frac{1}{a+2}+\frac{1}{b+2}\right)\)(1)
Tương tự ta có: \(\frac{1}{\sqrt{2\left(b^2+c^2\right)}+4}\le\frac{1}{4}\left(\frac{1}{b+2}+\frac{1}{c+2}\right)\)(2)
\(\frac{1}{\sqrt{2\left(c^2+a^2\right)}+4}\le\frac{1}{4}\left(\frac{1}{c+2}+\frac{1}{a+2}\right)\)(3)
Cộng từng vế của các bất đẳng thức (1), (2), (3), ta được:
\(P\le\frac{1}{2}\left(\frac{1}{a+2}+\frac{1}{b+2}+\frac{1}{c+2}\right)=\frac{1}{2}\)(theo (***))
Đẳng thức xảy ra khi \(a=b=c\)
Ta có: \(\left(a-b\right)^2\ge0\Leftrightarrow a^2-2ab+b^2\ge0\Leftrightarrow a^2+b^2\ge2ab\)
\(\Rightarrow\orbr{\begin{cases}a^2+2ab+b^2\ge4ab\\2\left(a^2+b^2\right)\ge a^2+2ab+b^2\end{cases}\Leftrightarrow\orbr{\begin{cases}a^2+2ab+b^2\ge4ab\\2\left(a^2+b^2\right)\ge a^2+2ab+b^2\end{cases}}}\)
\(\Leftrightarrow\orbr{\begin{cases}\left(a+b\right)^2\ge4ab\left(1\right)\\\left(a+b\right)^2\le2\left(a^2+b^2\right)\left(2\right)\end{cases}}\)
Theo đề bài:
\(a+b+3ab=1\)
\(\Leftrightarrow4\left(a+b\right)+12ab=4\)
\(\Leftrightarrow4\left(a+b\right)+3\left(a+b\right)^2\ge4\left(theo\left(1\right)\right)\)
\(\Leftrightarrow3\left(a+b\right)^2+4\left(a+b\right)-4\ge0\)
\(\Leftrightarrow\left(a+b+2\right)\left[3\left(a+b\right)-2\right]\ge0\)
\(\Leftrightarrow3\left(a+b\right)-2\ge0\left(a,b>0\Rightarrow a+b+2>0\right)\)
\(\Leftrightarrow a+b\ge\frac{2}{3}\)
`\(\Rightarrow2\left(a^2+b^2\right)\ge\left(a+b\right)^2\ge\frac{4}{9}\left(theo\left(2\right)\right)\)
Áp dụng các kết quả trên, ta có:
\(\left(\sqrt{1-a^2}+\sqrt{1-b^2}\right)^2\le2\left(1-a^2+1-b^2\right)\)\(=4-2\left(a^2+b^2\right)\le4-\frac{4}{9}=\frac{32}{9}\)
\(\Rightarrow\sqrt{1-a^2}+\sqrt{1-b^2}\le\frac{4\sqrt{2}}{3}\)
Ta có: \(\frac{3ab}{a+b}=\frac{1-\left(a+b\right)}{a+b}=\frac{1}{a+b}-1\le\frac{1}{\frac{2}{3}}-1=\frac{1}{2}\)
\(\Rightarrow A\le\frac{4\sqrt{2}}{3}+\frac{1}{2}\)
Dấu '=' xảy ra <=> \(\hept{\begin{cases}a=b\\a+b+3ab=1\end{cases}\Leftrightarrow\hept{\begin{cases}a=b\\3a^2+2a-1=0\end{cases}\Leftrightarrow}a=b=\frac{1}{3}\left(a,b>0\right)}\)
Vậy max A là \(\frac{4\sqrt{2}}{3}+\frac{1}{2}\Leftrightarrow a=b=\frac{1}{3}\)
\(\sqrt{2a^2+ab+2b^2}=\sqrt{\dfrac{3}{2}\left(a^2+b^2\right)+\dfrac{1}{2}\left(a+b\right)^2}\ge\sqrt{\dfrac{3}{4}\left(a+b\right)^2+\dfrac{1}{2}\left(a+b\right)^2}=\dfrac{\sqrt{5}}{2}\left(a+b\right)\)
Tương tự:
\(\sqrt{2b^2+bc+2c^2}\ge\dfrac{\sqrt{5}}{2}\left(b+c\right)\) ; \(\sqrt{2c^2+ca+2a^2}\ge\dfrac{\sqrt{5}}{2}\left(c+a\right)\)
Cộng vế với vế:
\(P\ge\sqrt{5}\left(a+b+c\right)\ge\dfrac{\sqrt{5}}{3}\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^3=\dfrac{\sqrt{5}}{3}\)
Dấu "=" xảy ra khi \(a=b=c=\dfrac{1}{9}\)
\(6a+3b+2c=abc\Leftrightarrow\dfrac{2}{ab}+\dfrac{3}{ac}+\dfrac{6}{bc}=1\)
Đặt \(\left(\dfrac{1}{a};\dfrac{2}{b};\dfrac{3}{c}\right)=\left(x;y;z\right)\Rightarrow xy+yz+zx=1\)
\(Q=\dfrac{1}{\sqrt{\dfrac{1}{x^2}+1}}+\dfrac{2}{\sqrt{\dfrac{4}{y^2}+4}}+\dfrac{3}{\sqrt{\dfrac{9}{z^2}+9}}=\dfrac{x}{\sqrt{x^2+1}}+\dfrac{y}{\sqrt{y^2+1}}+\dfrac{z}{\sqrt{z^2+1}}\)
\(Q=\dfrac{x}{\sqrt{x^2+xy+yz+zx}}+\dfrac{y}{\sqrt{y^2+xy+yz+zx}}+\dfrac{z}{\sqrt{z^2+xy+yz+zx}}\)
\(Q=\dfrac{x}{\sqrt{\left(x+y\right)\left(x+z\right)}}+\dfrac{y}{\sqrt{\left(x+y\right)\left(y+z\right)}}+\dfrac{z}{\sqrt{\left(x+z\right)\left(y+z\right)}}\)
\(Q\le\dfrac{1}{2}\left(\dfrac{x}{x+y}+\dfrac{x}{x+z}+\dfrac{y}{x+y}+\dfrac{y}{y+z}+\dfrac{z}{x+z}+\dfrac{z}{y+z}\right)=\dfrac{3}{2}\)
\(Q_{max}=\dfrac{3}{2}\) khi \(x=y=z=\dfrac{1}{\sqrt{3}}\) hay \(\left(a;b;c\right)=\left(\sqrt{3};2\sqrt{3};3\sqrt{3}\right)\)
\(Q\le\sqrt{3\left(a+b+b+c+c+a\right)}=\sqrt{6\left(a+b+c\right)}\le\sqrt{6.\sqrt{3\left(a^2+b^2+c^2\right)}}=\sqrt{6\sqrt{3}}\)
Dấu "=" xảy ra khi \(a=b=c=\dfrac{1}{\sqrt{3}}\)
Lại có:
\(a^2+b^2+c^2\le1\Rightarrow0\le a;b;c\le1\)
\(\Leftrightarrow a\left(a-1\right)+b\left(b-1\right)+c\left(c-1\right)\le0\)
\(\Leftrightarrow a+b+c\ge a^2+b^2+c^2=1\)
Do đó:
\(Q^2=2\left(a+b+c\right)+2\sqrt{a^2+ab+bc+ca}+2\sqrt{b^2+ab+bc+ca}+2\sqrt{c^2+ab+bc+ca}\)
\(Q^2\ge2\left(a+b+c\right)+2\sqrt{a^2}+2\sqrt{b^2}+2\sqrt{c^2}\)
\(Q^2\ge4\left(a+b+c\right)\ge4\)
\(\Rightarrow Q\ge2\)
Dấu "=" xảy ra khi \(\left(a;b;c\right)=\left(0;0;1\right)\) và hoán vị