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Bạn tham khảo, số liệu chỉ khác nhau đúng 1 chút xíu còn cách làm tương tự:
cho a,b,c dương thỏa mãn \(\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{c^2+a^2}=\sqrt{2011}\).CMR: \(\dfrac{a^2}{b+c}+\dfrac{b^2... - Hoc24
\(\left(a^2+\dfrac{b^2}{3}+\dfrac{c^2}{5}\right)\left(1+3+5\right)\ge\left(a+b+c\right)^2\)
\(\Rightarrow3\sqrt{a^2+\dfrac{b^2}{3}+\dfrac{c^2}{5}}\ge a+b+c\)
\(\Rightarrow P\ge\dfrac{2}{3}\left(a+b+c\right)+3\sqrt{\dfrac{1}{a}+\dfrac{3^2}{b}+\dfrac{5^2}{c}}\)
\(\Rightarrow P\ge\dfrac{2}{3}\left(a+b+c\right)+3\sqrt{\dfrac{\left(1+3+5\right)^2}{a+b+c}}=\dfrac{2}{3}\left(a+b+c\right)+\dfrac{27}{\sqrt{a+b+c}}\)
\(\Rightarrow P\ge\dfrac{1}{2}\left(a+b+c\right)+\dfrac{27}{2\sqrt{a+b+c}}+\dfrac{27}{2\sqrt{a+b+c}}+\dfrac{1}{6}\left(a+b+c\right)\)
\(\Rightarrow P\ge3\sqrt[3]{\dfrac{27^2\left(a+b+c\right)}{2^3\left(a+b+c\right)}}+\dfrac{1}{6}.9=15\)
Dấu "=" xảy ra khi \(\left(a;b;c\right)=\left(1;3;5\right)\)
a: \(=6\sqrt{a}+\dfrac{1}{3}\sqrt{a}-3\sqrt{a}+\sqrt{7}=\dfrac{10}{3}\sqrt{a}+\sqrt{7}\)
b: \(=5a\cdot5b\sqrt{ab}+\sqrt{3}\cdot2\sqrt{3}\cdot ab\sqrt{ab}+9ab\cdot3\sqrt{ab}-5b\cdot9a\sqrt{ab}\)
\(=25ab\sqrt{ab}+12ab\sqrt{ab}+27ab\sqrt{ab}-45ab\sqrt{ab}\)
\(=19ab\sqrt{ab}\)
c: \(=\dfrac{\sqrt{ab}}{b}+\sqrt{ab}-\dfrac{a}{b}\cdot\dfrac{\sqrt{b}}{\sqrt{a}}\)
\(=\sqrt{ab}\left(\dfrac{1}{b}+1\right)-\dfrac{\sqrt{a}}{\sqrt{b}}\)
\(=\sqrt{ab}\)
d: \(=11\sqrt{5a}-5\sqrt{5a}+2\sqrt{5a}-12\sqrt{5a}+9\sqrt{a}\)
\(=-4\sqrt{5a}+9\sqrt{a}\)
\(1,\text{Áp dụng Mincopxki: }\\ Q\ge\sqrt{\left(a+\dfrac{1}{a}\right)^2+\left(b+\dfrac{1}{b}\right)^2}\ge\sqrt{2^2+2^2}=\sqrt{8}=2\sqrt{2}\\ \text{Dấu }"="\Leftrightarrow a=b\)
\(2,\text{Áp dụng BĐT Cauchy-Schwarz: }\\ P\ge\dfrac{9}{a^2+b^2+c^2+2ab+2bc+2ca}=\dfrac{9}{\left(a+b+c\right)^2}\ge\dfrac{9}{1}=9\\ \text{Dấu }"="\Leftrightarrow a=b=c=\dfrac{1}{3}\)
Bạn tham khảo:
Cho \(a,b,c>\dfrac{25}{4}.\)Tìm GTNN của \(Q=\dfrac{a}{2\sqrt{b}-5}+\dfrac{b}{2\sqrt{c}-5}+\dfrac{c}{2\sqrt{a}-5}\) - Hoc24
2:
\(VT=\dfrac{a^2b}{a-b}\cdot\dfrac{2\sqrt{2}\left(a-b\right)}{5\sqrt{3}\cdot a^2\sqrt{b}}=\dfrac{2}{15}\cdot\sqrt{6b}=VP\)
1: \(=9\sqrt{ab}+\dfrac{7\sqrt{ab}}{b}-\dfrac{5\sqrt{ab}}{a}-3\sqrt{ab}=\)6căn ab+căn ab(7/b-5/a)
=căn ab(6+7/b-5/a)
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 $a=b=c=d=1$
\(A=\dfrac{a^2}{a\sqrt{a^2+9bc}}+\dfrac{b^2}{b\sqrt{b^2+9ca}}+\dfrac{c^2}{c\sqrt{c^2+9ab}}\)
\(A\ge\dfrac{\left(a+b+c\right)^2}{a\sqrt{a^2+9bc}+b\sqrt{b^2+9ca}+c\sqrt{c^2+9ab}}\)
Áp dụng Bunhiacopxki:
\(\sqrt{a}.\sqrt{a^3+9abc}+\sqrt{b}.\sqrt{b^3+9abc}+\sqrt{c}.\sqrt{c^3+9abc}\le\sqrt{\left(a+b+c\right)\left(a^3+b^3+c^3+27abc\right)}\)
\(\Rightarrow A\ge\dfrac{\left(a+b+c\right)^2}{\sqrt{\left(a+b+c\right)\left(a^3+b^3+c^3+27abc\right)}}=\sqrt{\dfrac{\left(a+b+c\right)^3}{a^3+b^3+c^3+27abc}}\) (1)
Ta có:
\(\left(a+b+c\right)^3=a^3+b^3+c^3+3\left(a^2b+b^2c+c^2a+ab^2+bc^2+ca^2\right)+6abc\)
\(\dfrac{1}{10}\left(a^3+b^3+c^3\right)\ge\dfrac{3}{10}abc\)
\(a^2b+b^2c+c^2a+ab^2+bc^2+ca^2\ge6\sqrt[6]{a^6b^6c^6}=6abc\)
\(\Rightarrow\left(a+b+c\right)^3\ge\dfrac{9}{10}\left(a^3+b^3+c^3\right)+\dfrac{3}{10}abc+18abc+6abc\)
\(\Rightarrow\left(a+b+c\right)^3\ge\dfrac{9}{10}\left(a^3+b^3+c^3+27abc\right)\) (2)
(1);(2) \(\Rightarrow A\ge\sqrt{\dfrac{\dfrac{9}{10}\left(a^3+b^3+c^3+27abc\right)}{a^3+b^3+c^3+27abc}}=\dfrac{3\sqrt{10}}{10}\)
Dấu "=" xảy ra khi \(a=b=c\)