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\(\sum\dfrac{a}{\sqrt{ab+b^2}}=\sum\dfrac{a\sqrt{2}}{\sqrt{2b\left(a+b\right)}}\ge\sum\dfrac{2\sqrt{2}a}{2b+a+b}=2\sqrt{2}\sum\dfrac{a}{a+3b}\)
\(=2\sqrt{2}\sum\dfrac{a^2}{a^2+3ab}\ge\dfrac{2\sqrt{2}\left(a+b+c\right)^2}{a^2+b^2+c^2+3\left(ab+bc+ca\right)}\)
\(=\dfrac{2\sqrt{2}\left(a+b+c\right)^2}{\left(a+b+c\right)^2+ab+bc+ca}\ge\dfrac{2\sqrt{2}\left(a+b+c\right)^2}{\left(a+b+c\right)^2+\dfrac{1}{3}\left(a+b+c\right)^2}=\dfrac{3\sqrt{2}}{2}\)
Lời giải:
Theo hệ quả quen thuộc của bđt AM-GM:
$(a+b+c)^2\leq 3(a^2+b^2+c^2)\leq 9$
$\Rightarrow a+b+c\leq 3$ (đpcm)
Từ đây ta có:
\(E\leq \frac{a}{\sqrt[3]{(a+b+c)a+bc}}+\frac{b}{\sqrt[3]{(a+b+c)b+ac}}+\frac{c}{\sqrt[3]{c(a+b+c)+ab}}\)
\(=\frac{a}{\sqrt[3]{(a+b)(a+c)}}+\frac{b}{\sqrt[3]{(b+c)(b+a)}}+\frac{c}{\sqrt[3]{(c+a)(c+b)}}\)
\(\leq \frac{\sqrt[3]{2}}{3}(\frac{a}{2}+\frac{a}{a+b}+\frac{a}{a+c})+\frac{\sqrt[3]{2}}{3}(\frac{b}{2}+\frac{b}{b+a}+\frac{b}{b+c})+\frac{\sqrt[3]{2}}{3}(\frac{c}{2}+\frac{c}{c+a}+\frac{c}{c+b})\)
\(=\frac{\sqrt[3]{2}(a+b+c)}{6}+\frac{\sqrt[3]{2}}{3}(\frac{a+b}{a+b}+\frac{b+c}{b+c}+\frac{c+a}{c+a})\leq \frac{3\sqrt[3]{2}}{2}\)
Vậy.................
\(3\ge a^2+b^2+c^2\ge\dfrac{1}{3}\left(a+b+c\right)^2\Rightarrow a+b+c\le3\)
\(\Rightarrow\dfrac{a}{\sqrt[3]{3a+bc}}\le\dfrac{a}{\sqrt[3]{a\left(a+b+c\right)+bc}}=\sqrt[3]{2}.\sqrt[3]{\dfrac{a}{a+b}.\dfrac{a}{a+c}.\dfrac{a}{2}}\le\dfrac{\sqrt[3]{2}}{3}\left(\dfrac{a}{a+b}+\dfrac{a}{a+c}+\dfrac{a}{2}\right)\)
Cộng vế và rút gọn:
\(E\le\dfrac{\sqrt[3]{2}}{3}\left(\dfrac{a}{a+b}+\dfrac{b}{a+b}+\dfrac{a}{a+c}+\dfrac{c}{a+c}+\dfrac{b}{b+c}+\dfrac{c}{b+c}+\dfrac{a+b+c}{2}\right)\)
\(E\le\dfrac{\sqrt[3]{2}}{3}\left(3+\dfrac{3}{2}\right)=\dfrac{3\sqrt[3]{2}}{2}\)
Bunhiacopxki:
\(\left(b+a+a\right)\left(b+c+\dfrac{c^2}{a}\right)\ge\left(b+\sqrt{ca}+c\right)^2\)
\(\Rightarrow\dfrac{2a^2+ab}{\left(b+\sqrt{ca}+c\right)^2}\ge\dfrac{2a^2+ab}{\left(2a+b\right)\left(b+c+\dfrac{c^2}{a}\right)}=\dfrac{a^2}{c^2+ab+bc}\)
Tương tự:
\(\dfrac{2b^2+bc}{\left(c+\sqrt{ca}+a\right)^2}\ge\dfrac{b^2}{a^2+ab+bc}\)
\(\dfrac{2c^2+ca}{\left(a+\sqrt{bc}+b\right)^2}\ge\dfrac{c^2}{b^2+ac+bc}\)
\(\Rightarrow P\ge\dfrac{a^2}{c^2+ab+ac}+\dfrac{b^2}{a^2+ab+bc}+\dfrac{c^2}{b^2+ac+bc}\)
\(\Rightarrow P\ge\dfrac{\left(a+b+c\right)^2}{a^2+b^2+c^2+2ab+2bc+2ca}=1\)
Dấu "=" xảy ra khi \(a=b=c\)
Áp dụng bất đẳng thức cộng mẫu số
\(\Rightarrow\dfrac{a^2}{a+\sqrt[3]{bc}}+\dfrac{b^2}{b+\sqrt[3]{ca}}+\dfrac{c^2}{c+\sqrt[3]{ab}}\ge\dfrac{\left(a+b+c\right)^2}{a+b+c+\sqrt[3]{bc}+\sqrt[3]{ca}+\sqrt[3]{ab}}\)
\(\Rightarrow\dfrac{a^2}{a+\sqrt[3]{bc}}+\dfrac{b^2}{b+\sqrt[3]{ca}}+\dfrac{c^2}{c+\sqrt[3]{ab}}\ge\dfrac{9}{3+\sqrt[3]{bc}+\sqrt[3]{ca}+\sqrt[3]{ab}}\)
Chứng minh rằng \(\dfrac{9}{3+\sqrt[3]{bc}+\sqrt[3]{ca}+\sqrt[3]{ab}}\ge\dfrac{3}{2}\)
\(\Leftrightarrow18\ge3\left(3+\sqrt[3]{bc}+\sqrt[3]{ca}+\sqrt[3]{ab}\right)\)
\(\Leftrightarrow18\ge9+3\sqrt[3]{bc}+3\sqrt[3]{ca}+3\sqrt[3]{ab}\)
\(\Leftrightarrow9\ge3\sqrt[3]{ab}+3\sqrt[3]{bc}+3\sqrt[3]{ca}\)
Áp dụng bất đẳng thức Cauchy cho 3 bộ số thực không âm
\(\Rightarrow\left\{{}\begin{matrix}a+b+1\ge3\sqrt[3]{ab}\\b+c+1\ge3\sqrt[3]{bc}\\c+a+1\ge3\sqrt[3]{ca}\end{matrix}\right.\)
\(\Rightarrow2\left(a+b+c\right)+3\ge3\sqrt[3]{ab}+3\sqrt[3]{bc}+3\sqrt[3]{ca}\)
\(\Rightarrow9\ge3\sqrt[3]{ab}+3\sqrt[3]{bc}+3\sqrt[3]{ca}\) ( đpcm )
Vì \(\dfrac{9}{3+\sqrt[3]{bc}+\sqrt[3]{ca}+\sqrt[3]{ab}}\ge\dfrac{3}{2}\)
Mà \(\dfrac{a^2}{a+\sqrt[3]{bc}}+\dfrac{b^2}{b+\sqrt[3]{ca}}+\dfrac{c^2}{c+\sqrt[3]{ab}}\ge\dfrac{9}{3+\sqrt[3]{bc}+\sqrt[3]{ca}+\sqrt[3]{ab}}\)
\(\Rightarrow\dfrac{a^2}{a+\sqrt[3]{bc}}+\dfrac{b^2}{b+\sqrt[3]{ca}}+\dfrac{c^2}{c+\sqrt[3]{ab}}\ge\dfrac{3}{2}\)( đpcm )
Áp dụng BĐT AM-GM và Cauchy-Schwarz ta có:
\(\sum\frac{a^2}{a+\sqrt[3]{bc}}\geq\sum\frac{a^2}{a+\frac{b+c+1}{3}}=\sum\frac{9a^2}{3(3a+b+c)+a+b+c}\)
\(=\sum\frac{9a^2}{10a+4b+4c}\geq\frac{9(a+b+c)^2}{(10a+4b+4c)}=\frac{9(a+b+c)^2}{18(a+b+c)}=\frac{3}{2}\)
\(\Leftrightarrow\dfrac{a}{\sqrt{4b^2+bc+4c^2}}+\dfrac{b}{\sqrt{4c^2+ca+4a^2}}+\dfrac{c}{\sqrt{4a^2+ab+4b^2}}\ge1\)
Ta có:
\(\sum\left(\dfrac{a}{\sqrt{4b^2+bc+4c^2}}\right)^2\sum a\left(4b^2+bc+4c^2\right)\ge\left(a+b+c\right)^3\)
Nên ta chỉ cần chứng minh:
\(\dfrac{\left(a+b+c\right)^3}{a\left(4b^2+bc+4c^2\right)+b\left(4c^2+ac+4a^2\right)+c\left(4a^2+ab+4b^2\right)}\ge1\)
\(\Leftrightarrow\dfrac{\left(a+b+c\right)^3}{4a\left(b^2+c^2\right)+4b\left(c^2+a^2\right)+4c\left(a^2+b^2\right)+3abc}\ge1\)
\(\Leftrightarrow a^3+b^3+c^3+3abc\ge ab\left(a+b\right)+bc\left(b+c\right)+ca\left(c+a\right)\) (đúng theo Schur bậc 3)
Đề bài sai
Đề đúng: \(\dfrac{1}{\sqrt{a}+2\sqrt{b}+3}+\dfrac{1}{\sqrt{b}+2\sqrt{c}+3}+\dfrac{1}{\sqrt{c}+2\sqrt{a}+3}\le\dfrac{1}{2}\)
\(\sqrt{\dfrac{ab+2c^2}{1+ab-c^2}}=\sqrt{\dfrac{ab+2c^2}{a^2+b^2+ab}}\)\(=\dfrac{ab+2c^2}{\sqrt{\left(a^2+b^2+ab\right)\left(ab+c^2+c^2\right)}}\)\(\ge\dfrac{2\left(ab+2c^2\right)}{a^2+b^2+2ab+2c^2}\)\(\ge\dfrac{2\left(ab+2c^2\right)}{2\left(a^2+b^2\right)+2c^2}\)\(=\dfrac{ab+2c^2}{a^2+b^2+c^2}\)
\(\Rightarrow\sqrt{\dfrac{ab+2c^2}{1+ab-c^2}}\ge ab+2c^2\)
Tương tự: \(\sqrt{\dfrac{bc+2a^2}{1+bc-a^2}}\ge bc+2a^2\); \(\sqrt{\dfrac{ac+2b^2}{1+ac-b^2}}\ge ac+2b^2\)
Cộng vế với vế \(\Rightarrow VT\ge2a^2+2b^2+2c^2+ab+bc+ac=2+ab+bc+ac\)
Dấu = xảy ra khi \(a=b=c=\dfrac{1}{\sqrt{3}}\)
Áp dụng BĐT Cosi:
\(\dfrac{a}{\sqrt{b^2+ab}}=\dfrac{a\sqrt{2}}{\sqrt{2\left(b^2+ab\right)}}=\dfrac{a\sqrt{2}}{\sqrt{2b\left(a+b\right)}}\ge\dfrac{a\sqrt{2}}{\dfrac{2b+a+b}{2}}=\dfrac{2\sqrt{2}a}{a+3b}\)
Cmtt: \(\dfrac{b}{\sqrt{c^2+bc}}\ge\dfrac{2\sqrt{2}b}{b+3c};\dfrac{c}{\sqrt{a^2+ca}}\ge\dfrac{2\sqrt{2}c}{c+3a}\)
\(\Leftrightarrow P\ge2\sqrt{2}\left(\dfrac{a}{a+3b}+\dfrac{b}{b+3c}+\dfrac{c}{c+3a}\right)\\ \Leftrightarrow\dfrac{P}{\sqrt{2}}\ge2\left(\dfrac{a}{a+3b}+\dfrac{b}{b+3c}+\dfrac{c}{c+3a}\right)\\ \Leftrightarrow\dfrac{P}{\sqrt{2}}\ge\dfrac{2\left(a+b+c\right)^2}{a^2+b^2+c^2+3\left(ab+bc+ca\right)}\ge\dfrac{2\left(a+b+c\right)^2}{\left(a+b+c\right)^2+\dfrac{1}{3}\left(a+b+c\right)^2}\\ \Leftrightarrow\dfrac{P}{\sqrt{2}}\ge\dfrac{2}{\dfrac{4}{3}}=\dfrac{3}{2}\\ \Leftrightarrow P\ge\dfrac{3\sqrt{2}}{2}\)
Dấu \("="\Leftrightarrow a=b=c\)
từ dòng thứ 4 lm sao suy ra dòng thứ 5 thế ạ