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Dự đoán điểm rơi xảy ra tại \(\left(a;b;c\right)=\left(3;2;4\right)\)
Đơn giản là kiên nhẫn tính toán và tách biểu thức:
\(D=13\left(\dfrac{a}{18}+\dfrac{c}{24}\right)+13\left(\dfrac{b}{24}+\dfrac{c}{48}\right)+\left(\dfrac{a}{9}+\dfrac{b}{6}+\dfrac{2}{ab}\right)+\left(\dfrac{a}{18}+\dfrac{c}{24}+\dfrac{2}{ac}\right)+\left(\dfrac{b}{8}+\dfrac{c}{16}+\dfrac{2}{bc}\right)+\left(\dfrac{a}{9}+\dfrac{b}{6}+\dfrac{c}{12}+\dfrac{8}{abc}\right)\)
Sau đó Cô-si cho từng ngoặc là được
Đặt \(\left(\sqrt{a};\sqrt{b};\sqrt{c}\right)=\left(x;y;z\right)\Rightarrow x+y+z=1\)
BĐT trở thành: \(\dfrac{xy}{\sqrt{x^2+y^2+2z^2}}+\dfrac{yz}{\sqrt{y^2+z^2+2x^2}}+\dfrac{zx}{\sqrt{x^2+z^2+2y^2}}\le\dfrac{1}{2}\)
Ta có:
\(x^2+z^2+y^2+z^2\ge\dfrac{1}{2}\left(x+z\right)^2+\dfrac{1}{2}\left(y+z\right)^2\ge\left(x+z\right)\left(y+z\right)\)
\(\Rightarrow\dfrac{xy}{\sqrt{x^2+y^2+2z^2}}\le\dfrac{xy}{\sqrt{\left(x+z\right)\left(y+z\right)}}\le\dfrac{1}{2}\left(\dfrac{xy}{x+z}+\dfrac{xy}{y+z}\right)\)
Tương tự: \(\dfrac{yz}{\sqrt{y^2+z^2+2x^2}}\le\dfrac{1}{2}\left(\dfrac{yz}{x+y}+\dfrac{yz}{x+z}\right)\)
\(\dfrac{zx}{\sqrt{z^2+x^2+2y^2}}\le\dfrac{1}{2}\left(\dfrac{zx}{x+y}+\dfrac{zx}{y+z}\right)\)
Cộng vế với vế:
\(VT\le\dfrac{1}{2}\left(\dfrac{zx+yz}{x+y}+\dfrac{xy+zx}{y+z}+\dfrac{yz+xy}{z+x}\right)=\dfrac{1}{2}\left(x+y+z\right)=\dfrac{1}{2}\) (đpcm)
Dấu "=" xảy ra khi \(x=y=z\) hay \(a=b=c\)
Áp dụng bđt AM - GM:
\(P=3a+3b-1+\left[\left(a+1\right)+b+\dfrac{c^3}{b\left(a+1\right)}\right]\ge3a+3b-1+3c=3.5-1=14\).
Đẳng thức xảy ra khi a = 1; b = 2; c = 2.
Vậy Min P = 14 khi a = 1; b = 2; c = 2.
Từ \(a^2b^2+b^2c^2+c^2a^2\ge a^2b^2c^2\)\(\Rightarrow\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}=1\)
bài này tui làm rồi ở đây
Ta có: \(\dfrac{a^3}{a^2+2b^2}=a-\dfrac{2ab^2}{a^2+2b^2}\ge a-\dfrac{2ab^2}{3\sqrt[3]{a^2b^4}}=a-\dfrac{2}{3}\sqrt[3]{ab^2}\ge a-\dfrac{2}{9}\left(a+b+b\right)=a-\dfrac{2}{9}\left(a+2b\right)\) Chứng minh tương tự ta được:
\(\dfrac{b^3}{b^2+2c^2}\ge b-\dfrac{2}{9}\left(b+2c\right);\dfrac{c^3}{c^2+2a^2}\ge c-\dfrac{2}{9}\left(c+2a\right)\)
\(\Rightarrow\dfrac{a^3}{a^2+2b^2}+\dfrac{b^3}{b^2+2c^2}+\dfrac{c^3}{c^2+2a^2}\ge a+b+c-\dfrac{2}{9}\left(a+2b+b+2c+c+2a\right)=a+b+c-\dfrac{2}{9}\left(3a+3b+3c\right)=\dfrac{1}{3}\left(a+b+c\right)\ge\dfrac{1}{3}\cdot3\sqrt[3]{abc}=1\)Dấu = xảy ra \(\Leftrightarrow a=b=c=1\)
Ta có:
\(\dfrac{a}{bc}+\dfrac{b}{ca}\ge2\sqrt{\dfrac{ab}{abc^2}}=\dfrac{2}{c}\)
Tương tự: \(\dfrac{a}{bc}+\dfrac{c}{ab}\ge\dfrac{2}{b}\) ; \(\dfrac{b}{ca}+\dfrac{c}{ab}\ge\dfrac{2}{a}\)
Cộng vế với vế: \(\Rightarrow\dfrac{a}{bc}+\dfrac{b}{ca}+\dfrac{c}{ab}\ge\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\)
\(\Rightarrow P\ge\dfrac{a^2+b^2+c^2}{2}+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\)
\(\Rightarrow P\ge\dfrac{1}{2}\left(a^2+\dfrac{1}{a}+\dfrac{1}{a}\right)+\dfrac{1}{2}\left(a^2+\dfrac{1}{b}+\dfrac{1}{b}\right)+\dfrac{1}{2}\left(c^2+\dfrac{1}{c}+\dfrac{1}{c}\right)\)
\(\Rightarrow P\ge\dfrac{1}{2}.3\sqrt[3]{\dfrac{a^2}{a^2}}+\dfrac{1}{2}.3\sqrt[3]{\dfrac{b^2}{b^2}}+\dfrac{1}{2}.3\sqrt[3]{\dfrac{c^2}{c^2}}=\dfrac{9}{2}\)
\(P_{min}=\dfrac{9}{2}\) khi \(a=b=c=1\)
a)Áp dụng BĐT AM-GM ta có:
\(\left\{{}\begin{matrix}x^2+y^2\ge2xy\\y^2+1\ge2y\end{matrix}\right.\)\(\Rightarrow x^2+2y^2+1\ge2xy+2y\)
\(\Rightarrow x^2+2y^2+3\ge2xy+2y+2\)
\(\Rightarrow\dfrac{1}{x^2+2y^2+3}\le\dfrac{1}{2\left(xy+y+1\right)}\Leftrightarrow\dfrac{2}{x^2+2y^2+3}\le\dfrac{1}{xy+y+1}\)
b)Áp dụng bổ đề trên ta có:
\(a^2+2b^2+3\ge2ab+2b+2\Rightarrow\dfrac{1}{a^2+2b^2+3}\le\dfrac{1}{2\left(ab+b+1\right)}\)
Tương tự cho 2 BĐT còn lại ta cũng có:
\(\dfrac{1}{b^2+2c^2+3}\le\dfrac{1}{2\left(bc+b+1\right)};\dfrac{1}{c^2+2a^2+3}\le\dfrac{1}{2\left(ac+c+1\right)}\)
Cộng theo vế 3 BĐT trên ta có:
\(Q\le\dfrac{1}{2\left(ab+b+1\right)}+\dfrac{1}{2\left(bc+b+1\right)}+\dfrac{1}{2\left(ac+c+1\right)}\)
\(=\dfrac{1}{2}\left(\dfrac{1}{ab+b+1}+\dfrac{1}{bc+b+1}+\dfrac{1}{ac+c+1}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{a}{ac+c+1}+\dfrac{ac}{ac+c+1}+\dfrac{1}{ac+c+1}\right)\left(abc=1\right)\)
\(=\dfrac{1}{2}\left(\dfrac{ac+c+1}{ac+c+1}\right)=\dfrac{1}{2}\)
Đẳng thức xảy ra khi \(x=y=z=1\)
\(a+b+c=2\Rightarrow ab+bc+ca\le\dfrac{\left(a+b+c\right)^2}{3}=\dfrac{4}{3}\)
\(P=\dfrac{7+2b}{1+a}+\dfrac{7+2c}{1+b}+\dfrac{7+2a}{1+c}\)
\(\ge\dfrac{\left(21+2\left(a+b+c\right)\right)^2}{\left(1+a\right)\left(7+2b\right)+\left(1+b\right)\left(7+2c\right)+\left(1+c\right)\left(7+2a\right)}\)
\(=\dfrac{25^2}{21+9\left(a+b+c\right)+2\left(ab+bc+ca\right)}\ge\dfrac{25^2}{21+9.2+\dfrac{2.4}{3}}=15\)
\("="\Leftrightarrow a=b=c=\dfrac{2}{3}\)