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Ta chứng minh bổ đề:
Với x,y,z dương thì:
\(8\left(x+y+z\right)\left(xy+yz+zx\right)\le9\left(x+y\right)\left(y+z\right)\left(z+x\right)\)
\(\Leftrightarrow x\left(y-z\right)^2+y\left(z-x\right)^2+z\left(x-y\right)^2\ge0\)(đúng)
Quay lại bài toán ta có:
\(A^{2020}=\left(\sqrt[2020]{\frac{a}{a+b}}+\sqrt[2020]{\frac{b}{b+c}}+\sqrt[2020]{\frac{c}{c+a}}\right)^{2020}\)
\(=\left(\sqrt[2020]{\frac{a\left(a+c\right)}{\left(a+b\right)\left(a+c\right)}}+\sqrt[2020]{\frac{b\left(b+a\right)}{\left(b+c\right)\left(b+a\right)}}+\sqrt[2020]{\frac{c\left(c+b\right)}{\left(c+a\right)\left(c+b\right)}}\right)^{2020}\)
\(\le\left(1+1+1\right)^{2018}.2.\left(a+b+c\right).\left(\frac{a}{\left(a+b\right)\left(a+c\right)}+\frac{b}{\left(b+c\right)\left(b+a\right)}+\frac{c}{\left(c+a\right)\left(c+b\right)}\right)\)
\(=3^{2018}.\frac{4\left(a+b+c\right)\left(ab+bc+ca\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)
\(\le3^{2018}.\frac{9\left(a+b\right)\left(b+c\right)\left(c+a\right)}{2\left(a+b\right)\left(b+c\right)\left(c+a\right)}=\frac{3^{2020}}{2}\)
\(\Rightarrow A\le\frac{3}{\sqrt[2020]{2}}\)
Từ giả thiết \(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}=1\Rightarrow xy+yz+xz=1\left(x=\dfrac{1}{a};y=\dfrac{1}{b};z=\dfrac{1}{c}\right)\)
\(A=\sum\dfrac{1}{\sqrt{1+a^2}}=\sum\dfrac{\dfrac{1}{a}}{\sqrt{\dfrac{1}{a^2}+1}}=\sum\dfrac{x}{\sqrt{x^2+1}}=\sum\dfrac{x}{\sqrt{x^2+xy+yz+xz}}=\sum\dfrac{x}{\sqrt{\left(x+y\right)\left(x+z\right)}}\le\dfrac{1}{2}\sum\dfrac{x}{x+y}+\dfrac{x}{x+z}=\dfrac{3}{2}\)
\(A\le\frac{1}{27}\left(\sqrt{4a+1}+\sqrt{4b+1}+\sqrt{4c+1}\right)^3\)
Mặt khác:
\(\sqrt{4a+1}+\sqrt{4b+1}+\sqrt{4c+1}\le\sqrt{3\left[4\left(a+b+c\right)+3\right]}=3\sqrt{5}\)
\(\Rightarrow A\le\frac{1}{27}\left(3\sqrt{5}\right)^3=5\sqrt{5}\)
Dấu "=" xảy ra khi \(a=b=c=1\)
Từ giả thiết ta có: \(1=a+b+c\ge3\sqrt[3]{abc}\Rightarrow abc\le\frac{1}{27}\)
Áp dụng BĐT AM - GM:
\(P=\frac{\sqrt{3}}{2}.\sqrt{\frac{4}{3}.a\left(a+bc\right)}+\frac{\sqrt{3}}{2}.\sqrt{\frac{4}{3}.b\left(b+ca\right)}+\frac{\sqrt{3}}{2}.\sqrt{\frac{4}{3}.c\left(c+ab\right)}+9\sqrt{abc}\)\(\le\frac{\sqrt{3}}{2}.\left(\frac{\frac{7}{3}a+bc+\frac{7}{3}b+ca+\frac{7}{3}c+ab}{2}\right)+9\sqrt{abc}\)
\(=\frac{\sqrt{3}}{2}.\left[\frac{\frac{7}{3}\left(a+b+c\right)+ab+bc+ca}{2}\right]+9\sqrt{abc}\)
\(=\frac{\sqrt{3}}{2}.\left(\frac{7}{6}+\frac{ab+bc+ca}{2}\right)+9\sqrt{abc}\)
Áp dụng BĐT quen thuộc \(xy+yz+zx\le\frac{\left(x+y+z\right)^2}{3}\)
Khi đó: \(P\le\frac{\sqrt{3}}{2}.\left(\frac{7}{6}+\frac{\frac{1}{3}}{2}\right)+9\sqrt{\frac{1}{27}}=\frac{5\sqrt{3}}{3}\)
\(\Rightarrow min_P=\frac{5\sqrt{3}}{3}\Leftrightarrow a=b=c=\frac{1}{3}\)
1/ a/dung bđt Cauchy - Schwarz dạng phân thức: \(\frac{a^2}{b+3c}+\frac{b^2}{c+3a}+\frac{c^2}{a+3b}\ge\frac{\left(a+b+c\right)^2}{4\left(a+b+c\right)}=\frac{a+b+c}{4}=\frac{3}{4}\)
2/ a/dung bđt bunhiacopxki :
\(S^2=\left(\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\right)^2\le\left(1^2+1^2+1^2\right)\left(a+b+b+c+c+a\right)=3\cdot2\left(a+b+c\right)=6\cdot6=36\)
=> \(S\le6\)
1.
\(6=\frac{\sqrt{2}^2}{x}+\frac{\sqrt{3}^2}{y}\ge\frac{\left(\sqrt{2}+\sqrt{3}\right)^2}{x+y}=\frac{5+2\sqrt{6}}{x+y}\)
\(\Rightarrow x+y\ge\frac{5+2\sqrt{6}}{6}\)
Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}\frac{x}{\sqrt{2}}=\frac{y}{\sqrt{3}}\\x+y=\frac{5+2\sqrt{6}}{6}\end{matrix}\right.\)
Bạn tự giải hệ tìm điểm rơi nếu thích, số xấu quá
2.
\(VT\ge\sqrt{\left(x+y+z\right)^2+\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2}\ge\sqrt{\left(x+y+z\right)^2+\frac{81}{\left(x+y+z\right)^2}}\)
Đặt \(x+y+z=t\Rightarrow0< t\le1\)
\(VT\ge\sqrt{t^2+\frac{81}{t^2}}=\sqrt{t^2+\frac{1}{t^2}+\frac{80}{t^2}}\ge\sqrt{2\sqrt{\frac{t^2}{t^2}}+\frac{80}{1^2}}=\sqrt{82}\)
Dấu "=" xảy ra khi \(x=y=z=\frac{1}{3}\)
3.
\(\frac{a^2}{b^5}+\frac{a^2}{b^5}+\frac{a^2}{b^5}+\frac{1}{a^3}+\frac{1}{a^3}\ge5\sqrt[5]{\frac{a^6}{b^{15}.a^6}}=\frac{5}{b^3}\)
Tương tự: \(\frac{3b^2}{c^5}+\frac{2}{b^3}\ge\frac{5}{a^3}\) ; \(\frac{3c^2}{d^5}+\frac{2}{c^3}\ge\frac{5}{d^3}\) ; \(\frac{3d^2}{a^5}+\frac{2}{d^2}\ge\frac{5}{a^3}\)
Cộng vế với vế và rút gọn ta được: \(3VT\ge3VP\)
Dấu "=" xảy ra khi và chỉ khi \(a=b=c=d=1\)
4.
ĐKXĐ: \(-2\le x\le2\)
\(y^2=\left(x+\sqrt{4-x^2}\right)^2\le2\left(x^2+4-x^2\right)=8\)
\(\Rightarrow y\le2\sqrt{2}\Rightarrow y_{max}=2\sqrt{2}\) khi \(x=\sqrt{2}\)
Mặt khác do \(\left\{{}\begin{matrix}x\ge-2\\\sqrt{4-x^2}\ge0\end{matrix}\right.\) \(\Rightarrow x+\sqrt{4-x^2}\ge-2\)
\(y_{min}=-2\) khi \(x=-2\)
we have that: \(\sqrt{a^4+b^2+c^2+1}=\sqrt{a^4-a^2+2}\)
and \(\dfrac{-a^2+11}{8}\le\sqrt{a^4-a^2+2}\le\sqrt{2}\) \(\left(a\in\left(0;1\right)\right)\)
Áp dụng BĐT Cauchy-Schwarz ta có:
\(A^2=\left(\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\right)\)
\(\le\left[\left(a+b\right)+\left(b+c\right)+\left(c+a\right)\right]\left(1+1+1\right)\)
\(=3\cdot2\left(a+b+c\right)=6\cdot4=24\)
\(\Rightarrow A^2\le24\Rightarrow A\le\sqrt{24}\)
Đẳng thức xảy ra khi \(a=b=c=\dfrac{4}{3}\)
hình như thiếu mũ 2 ở VP ở dòng đầu r`...