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Lời giải:
Bài 1:
Áp dụng BĐT Cô -si ta có:
\(a^3+1+1\geq 3\sqrt[3]{a^3}=3a\)
\(b^3+1+1\geq 3\sqrt[3]{b^3}=3b\)
Cộng theo vế:
\(a^3+b^3+4\geq 3(a+b)\)
\(\Leftrightarrow 6\geq 3(a+b)\Leftrightarrow a+b\leq 2\)
Vậy \((a+b)_{\max}=2\). Dấu bằng xảy ra khi \(a=b=1\)
Bài 2:
Áp dụng BĐT Cô- si ta có:
\(\frac{a^3}{b+c}+\frac{b+c}{4}+\frac{1}{2}\geq 3\sqrt[3]{\frac{a^3}{8}}=\frac{3}{2}a\)
\(\frac{b^3}{c+a}+\frac{c+a}{4}+\frac{1}{2}\geq 3\sqrt[3]{\frac{b^3}{8}}=\frac{3}{2}b\)
\(\frac{c^3}{a+b}+\frac{a+b}{4}+\frac{1}{2}\geq 3\sqrt[3]{\frac{c^3}{8}}=\frac{3}{2}c\)
Cộng theo vế:
\(T+\frac{1}{2}(a+b+c)+\frac{3}{2}\geq \frac{3}{2}(a+b+c)\)
\(\Leftrightarrow T\geq a+b+c-\frac{3}{2}\)
Theo BĐT Cô-si: \(a+b+c\geq 3\sqrt[3]{abc}=3\)
\(\Rightarrow T\geq 3-\frac{3}{2}=\frac{3}{2}\)
Vậy \(T_{\min}=\frac{3}{2}\Leftrightarrow a=b=c=1\)
Bài 3:
Điều kiện đề bài tương đương với:
\(a\leq 1; b+2a\leq 4; 2c+3b+6a\leq 18\)
Ta có:
\(A=2\left (\frac{1}{6a}+\frac{1}{3b}+\frac{1}{2c}\right)+\frac{1}{3}\left(\frac{1}{2a}+\frac{1}{b}\right)+\frac{1}{2a}\)
Áp dụng BĐT Bunhiacopxky:
\(\left(\frac{1}{6a}+\frac{1}{3b}+\frac{1}{2c}\right)(6a+3b+2c)\geq (1+1+1)^2\)
\(\Rightarrow \frac{1}{6a}+\frac{1}{3b}+\frac{1}{2c}\geq \frac{9}{6a+3b+2c}\geq \frac{9}{18}=\frac{1}{2}\) (1)
\(\left(\frac{1}{2a}+\frac{1}{b}\right)(2a+b)\geq (1+1)^2\)
\(\Rightarrow \frac{1}{2a}+\frac{1}{b}\geq \frac{4}{2a+b}\geq \frac{4}{4}=1\) (2)
\(\frac{1}{2a}\geq \frac{1}{2.1}=\frac{1}{2}\) (3)
Từ (1)(2)(3) suy ra \(A\geq 2.\frac{1}{2}+\frac{1}{3}.1+\frac{1}{2}=\frac{11}{6}\)
Dấu bằng xảy ra khi \(a=1; b=2; c=3\)
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 \(x=-2\)
áp dụng bất đẳng thứcxvaco \(\frac{a^2}{x}+\frac{b^2}{y}+\frac{c^2}{z}\ge\frac{\left(a+b+c\right)^2}{x+y+z}\)
suy ra P >= (a+b+c)^2/ 2 (a+b+c)=1/2
Dấu bằng xảy ra <=> \(\frac{a}{b+c}=\frac{b}{c+a}=\frac{c}{a+b}\)
\(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\Leftrightarrow\frac{a+b}{ab}\ge\frac{4}{a+b}\)
\(\Leftrightarrow\left(a+b\right)^2\ge4ab\Leftrightarrow\left(a-b\right)^2\ge0\) (luôn đúng)
a/ \(VT=\frac{1}{a+a+b+c}+\frac{1}{a+b+b+c}+\frac{1}{a+b+c+c}\le\frac{1}{16}\left(\frac{1}{a}+\frac{1}{a}+\frac{1}{b}+\frac{1}{b}+\frac{1}{a}+\frac{1}{b}+\frac{1}{b}+\frac{1}{c}+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{c}\right)\)
\(\Rightarrow VT\le\frac{1}{4}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=1\) (đpcm)
Dấu "=" xảy ra khi \(a=b=c=\frac{3}{4}\)
b/ \(VT\le\frac{ab}{4}\left(\frac{1}{a}+\frac{1}{b}\right)+\frac{bc}{4}\left(\frac{1}{b}+\frac{1}{c}\right)+\frac{ca}{4}\left(\frac{1}{c}+\frac{1}{a}\right)\)
\(VT\le\frac{a}{4}+\frac{b}{4}+\frac{b}{4}+\frac{c}{4}+\frac{c}{4}+\frac{a}{4}=\frac{a+b+c}{2}\)
Dấu "=" xảy ra khi \(a=b=c\)
Mấy cái dấu "=" anh tự xét.
Áp dụng BĐT AM-GM: \(VT=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge3\sqrt[3]{\frac{1}{abc}}=\frac{3}{\sqrt[3]{abc}}\ge\frac{3}{\frac{a+b+c}{3}}=\frac{9}{a+b+c}\)
a) Áp dụng: \(VT\ge\frac{\left(a+b+c\right)^2}{3}.\frac{9}{2\left(a+b+c\right)}=\frac{3}{2}\left(a+b+c\right)\)
b) \(P=3-\left(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\right)\le3-\frac{9}{x+y+z+3}=\frac{3}{4}\)
\(P=a-\frac{ab^2}{1+b^2}+b-\frac{bc^2}{1+c^2}+c-\frac{ca^2}{1+a^2}\)
\(\ge a-\frac{ab^2}{2b}+b-\frac{bc^2}{2c}+c-\frac{ca^2}{2c}\) (AM-GM)
\(\ge a-\frac{ab}{2}+b-\frac{bc}{2}+c-\frac{ac}{2}\ge\left(a+b+c\right)-\frac{\left(a+b+c\right)^2}{6}\ge3-\frac{3}{2}=\frac{3}{2}\)
Vay MinP=3/2 dau = xay ra khi a=b=c=1