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Câu 1
\(a+b\ge2\sqrt{ab}\Leftrightarrow ab\le\dfrac{\left(a+b\right)^2}{4}\\ \Leftrightarrow N=ab+\dfrac{1}{16ab}+\dfrac{15}{16ab}\ge2\sqrt{\dfrac{1}{16}}+\dfrac{15}{4\left(a+b\right)^2}\ge\dfrac{1}{2}+\dfrac{15}{4}=\dfrac{17}{4}\)
Dấu \("="\Leftrightarrow a=b=\dfrac{1}{2}\)
Câu 2:
\(P=a+\dfrac{1}{a}+2b+\dfrac{8}{b}+3c+\dfrac{27}{c}+4\left(a+b+c\right)\\ P\ge2\sqrt{1}+2\sqrt{16}+2\sqrt{81}+4\cdot6=2+8+18+4=32\)
Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}a=1\\b=2\\c=3\end{matrix}\right.\)
Câu 3: Cho a,b,c là các số thuộc đoạn [ -1;2 ] thõa mãn \(a^2+b^2+c^2=6.\) CMR : \(a+b+c>0\) - Hoc24
Theo giả thiết: \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\) \(\Leftrightarrow\) \(\dfrac{ab+bc+ca}{abc}=0\)
\(\Leftrightarrow\) ab+bc+ca = 0 \(\Rightarrow\) \(\left\{{}\begin{matrix}ab=-ac-bc\\bc=-ab-ac\\ac=-bc-ab\end{matrix}\right.\)
Xét \(\dfrac{a^2}{a^2+2bc}\) = \(\dfrac{a^2}{a^2+bc-ab-ac}\) = \(\dfrac{a^2}{a\left(a-c\right)-b\left(a-c\right)}\)
= \(\dfrac{a^2}{\left(a-b\right)\left(a-c\right)}\)
CMTT \(\dfrac{b^2}{b^2+2ac}\) = \(\dfrac{b^2}{\left(b-a\right)\left(b-c\right)}\)
\(\dfrac{c^2}{c^2+2ab}\) = \(\dfrac{c^2}{\left(c-a\right)\left(c-b\right)}\)
\(\Rightarrow\) p = \(\dfrac{a^2}{\left(a-b\right)\left(a-c\right)}\)+\(\dfrac{b^2}{\left(b-a\right)\left(b-c\right)}\)+\(\dfrac{c^2}{\left(c-a\right)\left(c-b\right)}\)
= \(\dfrac{a^2\left(b-c\right)-b^2\left(a-c\right)+c^2\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\) = 1
Lời giải:
Áp dụng BĐT Bunhiacopxky:
\(\left(a^2+\frac{1}{b^2}\right)(1+1)\geq (a+\frac{1}{b})^2\)
\(\Rightarrow \sqrt{a^2+\frac{1}{b^2}}\geq \frac{a+\frac{1}{b}}{\sqrt{2}}\)
Hoàn toàn tương tự với các phân thức còn lại và cộng theo vế:
\(\sqrt{a^2+\frac{1}{b^2}}+\sqrt{b^2+\frac{1}{c^2}}+\sqrt{c^2+\frac{1}{d^2}}+\sqrt{d^2+\frac{1}{a^2}}\geq \frac{1}{\sqrt{2}}(a+b+c+d+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d})\)
Mặt khác theo BĐT Cauchy:
\(a+\frac{1}{a}\geq 2; b+\frac{1}{b}\geq 2; c+\frac{1}{c}\geq 2; d+\frac{1}{d}\geq 2\)
\(\Rightarrow \text{VT}\geq \frac{1}{\sqrt{2}}.8=4\sqrt{2}\)
Vậy giá trị nhỏ nhất của biểu thức là $4\sqrt{2}$. Dấu bằng xảy ra khi $a=b=c=d=1$
\(H=\sqrt{a^2+\dfrac{1}{b^2}}+\sqrt{b^2+\dfrac{1}{c^2}}+\sqrt{c^2+\dfrac{1}{a^2}}\)
\(\ge\sqrt{\left(a+b+c\right)^2+\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2}\)
\(\ge\sqrt{\left(a+b+c\right)^2+\dfrac{81}{\left(a+b+c\right)^2}}\)
\(\ge\sqrt{\left(\dfrac{3}{2}\right)^2+\dfrac{81}{\left(\dfrac{3}{2}\right)^2}}=\dfrac{3\sqrt{17}}{2}\)
\(\left(a^2+\dfrac{b^2}{3}+\dfrac{c^2}{5}\right)\left(1+3+5\right)\ge\left(a+b+c\right)^2\)
\(\Rightarrow3\sqrt{a^2+\dfrac{b^2}{3}+\dfrac{c^2}{5}}\ge a+b+c\)
\(\Rightarrow P\ge\dfrac{2}{3}\left(a+b+c\right)+3\sqrt{\dfrac{1}{a}+\dfrac{3^2}{b}+\dfrac{5^2}{c}}\)
\(\Rightarrow P\ge\dfrac{2}{3}\left(a+b+c\right)+3\sqrt{\dfrac{\left(1+3+5\right)^2}{a+b+c}}=\dfrac{2}{3}\left(a+b+c\right)+\dfrac{27}{\sqrt{a+b+c}}\)
\(\Rightarrow P\ge\dfrac{1}{2}\left(a+b+c\right)+\dfrac{27}{2\sqrt{a+b+c}}+\dfrac{27}{2\sqrt{a+b+c}}+\dfrac{1}{6}\left(a+b+c\right)\)
\(\Rightarrow P\ge3\sqrt[3]{\dfrac{27^2\left(a+b+c\right)}{2^3\left(a+b+c\right)}}+\dfrac{1}{6}.9=15\)
Dấu "=" xảy ra khi \(\left(a;b;c\right)=\left(1;3;5\right)\)
\(1,\text{Áp dụng Mincopxki: }\\ Q\ge\sqrt{\left(a+\dfrac{1}{a}\right)^2+\left(b+\dfrac{1}{b}\right)^2}\ge\sqrt{2^2+2^2}=\sqrt{8}=2\sqrt{2}\\ \text{Dấu }"="\Leftrightarrow a=b\)
\(2,\text{Áp dụng BĐT Cauchy-Schwarz: }\\ P\ge\dfrac{9}{a^2+b^2+c^2+2ab+2bc+2ca}=\dfrac{9}{\left(a+b+c\right)^2}\ge\dfrac{9}{1}=9\\ \text{Dấu }"="\Leftrightarrow a=b=c=\dfrac{1}{3}\)