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Ta thấy: \(a+b\le1\Leftrightarrow\hept{\begin{cases}a\le1-b\\b\le1-a\end{cases}}\Leftrightarrow\hept{\begin{cases}1+a\le2-b\\1+b\le2-a\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}\frac{a}{1+b}\ge\frac{a}{2-a}\\\frac{b}{1+a}\ge\frac{b}{2-b}\end{cases}}\Rightarrow\frac{a}{1+b}+\frac{b}{1+a}\ge\frac{a}{2-a}+\frac{b}{2-b}\)
\(\Rightarrow S=\frac{a}{1+b}+\frac{b}{1+a}+\frac{1}{a+b}\ge\frac{a}{2-a}+\frac{b}{2-b}+\frac{1}{a+b}\)
\(=\frac{2}{2-a}-1+\frac{2}{2-b}-1+\frac{1}{a+b}=\frac{2}{2-a}+\frac{2}{2-b}+\frac{1}{a+b}-2\)
\(=2\left(\frac{1}{2-a}+\frac{1}{2-b}+\frac{1}{2\left(a+b\right)}-1\right)\)
Áp dụng bất đẳng thức sau: \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{9}{x+y+z}\)
\(\Rightarrow\frac{1}{2-a}+\frac{1}{2-b}+\frac{1}{2\left(a+b\right)}\ge\frac{9}{4-\left(a+b\right)+2\left(a+b\right)}=\frac{9}{4+a+b}\)
Lại có: \(a+b\le1\Rightarrow4+a+b\le5\Rightarrow\frac{9}{4+a+b}\ge\frac{9}{5}\)
\(\Rightarrow\frac{1}{2-a}+\frac{1}{2-b}+\frac{1}{2\left(a+b\right)}\ge\frac{9}{5}\Leftrightarrow2\left(\frac{1}{2-a}+\frac{1}{2-b}+\frac{1}{2\left(a+b\right)}-1\right)\ge\frac{8}{5}\)
\(\Rightarrow S\ge\frac{8}{5}.\)
Vậy \(Min_S=\frac{8}{5}.\)Dấu "=" xảy ra khi \(a=b=\frac{2}{5}.\)
\(A=\dfrac{2}{a^2+b^2}+\dfrac{35}{ab}+2ab\)
\(=2\left(\dfrac{1}{a^2+b^2}+\dfrac{1}{2ab}\right)+\dfrac{34}{ab}+\dfrac{17}{8}ab-\dfrac{1}{8}ab\)
\(\ge2.\dfrac{4}{a^2+b^2+2ab}+2\sqrt{\dfrac{34}{ab}.\dfrac{17}{8}ab}-\dfrac{1}{8}.\dfrac{\left(a+b\right)^2}{4}\)
\(\Leftrightarrow A\ge2.\dfrac{4}{\left(a+b\right)^2}+2.\dfrac{17}{2}-\dfrac{1}{8}.\dfrac{4^2}{4}\ge2.\dfrac{4}{4^2}+17-\dfrac{1}{2}\)
\(\Leftrightarrow A\ge\dfrac{1}{2}+17-\dfrac{1}{2}=17\)
Dấu "=" <=> a = b = 2
Lời giải:
Ta dự toán cực trị xảy ra tại \(a=b=2\). Công việc còn lại là phân tích hợp lý.
Áp dụng BĐT Bunhiacopxky:
\(\left(\frac{2}{a^2+b^2}+\frac{2}{2ab}\right)(a^2+b^2+2ab)\geq (\sqrt{2}+\sqrt{2})^2\)
\(\Leftrightarrow \frac{2}{a^2+b^2}+\frac{1}{ab}\geq \frac{8}{a^2+b^2+2ab}=\frac{8}{(a+b)^2}\)
Mà \(a+b\lè 4\Rightarrow \frac{2}{a^2+b^2}+\frac{1}{ab} \geq \frac{8}{(a+b)^2}\geq \frac{8}{4^2}=\frac{1}{2}(1)\)
Áp dụng BĐT AM-GM:
\(\frac{32}{ab}+2ab\geq 2\sqrt{32.2}=16(2)\)
Tiếp tục AM-GM: \(4\geq a+b\geq 2\sqrt{ab}\Rightarrow ab\leq 4\)
\(\Rightarrow \frac{2}{ab}\geq \frac{2}{4}=\frac{1}{2}(3)\)
Lấy \((1)+(2)+(3)\Rightarrow A\geq \frac{1}{2}+16+\frac{1}{2}=17\)
Vậy \(A_{\min}=17\Leftrightarrow a=b=2\)
\(x=\dfrac{1}{2}\cdot\left(\dfrac{a}{\sqrt{ab}}+\dfrac{b}{\sqrt{ab}}\right)=\dfrac{a+b}{2\sqrt{ab}}\)
\(2\sqrt{x^2}-1=2\cdot\dfrac{a+b}{2\sqrt{ab}}-1=\dfrac{a+b-\sqrt{ab}}{\sqrt{ab}}\)
\(x-\sqrt{x^2-1}=\dfrac{a+b}{2\sqrt{ab}}-\sqrt{\dfrac{a^2+2ab+b^2}{4ab}-1}\)
\(=\dfrac{a+b}{2\sqrt{ab}}-\dfrac{a-b}{2\sqrt{ab}}=\dfrac{2b}{2\sqrt{ab}}=\dfrac{\sqrt{b}}{\sqrt{a}}\)
\(G=\dfrac{a+b-\sqrt{ab}}{\sqrt{ab}}:\dfrac{\sqrt{b}}{\sqrt{a}}=\dfrac{a+b-\sqrt{ab}}{\sqrt{ab}}\cdot\dfrac{\sqrt{a}}{\sqrt{b}}\)
\(=\dfrac{a+b-\sqrt{ab}}{b}\)
a)
Q=a√a2−b2−(1+a√a2−b2):ba−√a2−b2=a√a2−b2−a2−(a2−b2)b√a2−b2=a√a2−b2−a2−a2+b2b√a2−b2=a−b√a2−
Áp dụng BĐT svacxơ, ta có
\(\frac{1}{x^2+xy}+\frac{1}{y^2+xy}\ge\frac{4}{x^2+y^2+2xy}=\frac{4}{\left(x+y\right)^2}\ge4\)
Dấu = xảy ra <=>x=y=1/2
^_^
\(M=\left(a^2+\frac{1}{16a^2}\right)+\left(b^2+\frac{1}{16b^2}\right)+\frac{15}{16}\left(\frac{1}{a^2}+\frac{1}{b^2}\right)\)
\(\ge2\sqrt{\frac{a^2}{16a^2}}+2\sqrt{\frac{b^2}{16b^2}}+\frac{15\left(\frac{1}{a}+\frac{1}{b}\right)^2}{32}\ge1+\frac{\frac{240}{\left(a+b\right)^2}}{32}\ge\frac{17}{2}\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(a=b=\frac{1}{2}\)
\(A=\frac{2}{a^2+b^2}+\frac{35}{ab}+2ab\)
\(=2\left(\frac{1}{a^2+b^2}+\frac{1}{2ab}\right)+\frac{34}{ab}+\frac{17}{8}ab-\frac{1}{8}ab\)
\(\ge2.\frac{4}{a^2+b^2+2ab}+2\sqrt{\frac{34}{ab}.\frac{17}{8}ab}-\frac{1}{8}.\frac{\left(a+b\right)^2}{4}\)
\(\Leftrightarrow A\ge2.\frac{4}{\left(a+b\right)^2}+2.\frac{17}{2}-\frac{1}{8}.\frac{4}{4^2}+17-\frac{1}{2}\)
\(\Leftrightarrow A\ge\frac{1}{2}+17-\frac{1}{2}=17\)
Dấu " = " xảy ra \(\Leftrightarrow a=b=2\)
Chúc bạn học tốt !!!
\(A=a^2+b^2+\dfrac{1}{a^2}+\dfrac{1}{b^2}\)
\(A=a^2+\dfrac{1}{16a^2}+b^2+\dfrac{1}{16b^2}+\dfrac{15}{16}\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}\right)\)
\(A\ge2\sqrt{a^2\cdot\dfrac{1}{16a^2}}+2\sqrt{b^2\cdot\dfrac{1}{16b^2}}+\dfrac{15}{16}\cdot2\cdot\sqrt{\dfrac{1}{a^2b^2}}\)
\(A\ge1+\dfrac{15}{8ab}\ge1+\dfrac{15}{2\left(a+b\right)^2}\ge\dfrac{17}{2}\)
"="<=>x=y=0,5