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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:
\(\sum\frac{a^2}{a+\sqrt[3]{bc}}\geq\sum\frac{a^2}{a+\frac{b+c+1}{3}}=\sum\frac{9a^2}{3(3a+b+c)+a+b+c}\)
\(=\sum\frac{9a^2}{10a+4b+4c}\geq\frac{9(a+b+c)^2}{(10a+4b+4c)}=\frac{9(a+b+c)^2}{18(a+b+c)}=\frac{3}{2}\)
Lời giải ở đây: https://hoc24.vn/hoi-dap/question/486195.html
Lời giải:
Theo BĐT Cauchy Schwarz:
\(ab+bc+ac=3abc\Rightarrow 3=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq \frac{9}{a+b+c}\)
\(\Rightarrow a+b+c\geq 3\)
Áp dụng BĐT AM-GM:
\(A=a-\frac{ca}{c+a^2}+b-\frac{ab}{a+b^2}+c-\frac{bc}{b+c^2}\)
\(=(a+b+c)-\left(\frac{ac}{c+a^2}+\frac{ab}{a+b^2}+\frac{bc}{b+c^2}\right)\)
\(\geq (a+b+c)-\left(\frac{ac}{2a\sqrt{c}}+\frac{ab}{2b\sqrt{a}}+\frac{bc}{2c\sqrt{b}}\right)\)
\(A\geq (a+b+c)-\frac{\sqrt{a}+\sqrt{b}+\sqrt{c}}{2}\)
Cũng theo BĐT AM-GM:
\(\sqrt{a}+\sqrt{b}+\sqrt{c}\leq \frac{a+1}{2}+\frac{b+1}{2}+\frac{c+1}{2}=\frac{a+b+c+1}{4}\)
\(\Rightarrow A\geq a+b+c-\frac{a+b+c+3}{4}=\frac{3}{4}(a+b+c)-\frac{3}{4}\geq \frac{3}{4}.3-\frac{3}{4}=\frac{3}{2}\)
Vậy \(A_{\min}=\frac{3}{2}\Leftrightarrow a=b=c=1\)
Lời giải:
Từ \(ab+bc+ac=1\Rightarrow a^2+ab+bc+ac=a^2+1\)
\(\Leftrightarrow (a+b)(a+c)=a^2+1\)
Tương tự: \(\left\{\begin{matrix} b^2+1=(b+c)(b+a)\\ c^2+1=(c+a)(c+b)\end{matrix}\right.\)
Khi đó:
\(A=\frac{(b^2+bc)(c^2+ca)(a^2+ab)}{\sqrt{(a^4+a^2)(b^4+b^2)(c^4+c^2)}}\) \(=\frac{b(b+c)c(c+a)a(a+b)}{\sqrt{a^2b^2c^2(a^2+1)(b^2+1)(c^2+1)}}\)
\(=\frac{abc(a+b)(b+c)(c+a)}{abc\sqrt{(a+b)(a+c)(b+c)(b+a)(c+a)(c+b)}}\) \(=\frac{abc(a+b)(b+c)(c+a)}{abc(a+b)(b+c)(c+a)}=1\)
Vậy \(A=1\)
Lời giải:
Do \(3=ab+bc+ac\) nên ta có:
\(P=\frac{a^3}{b^2+3}+\frac{b^3}{c^2+3}+\frac{c^3}{a^2+3}\)
\(=\frac{a^3}{b^2+ab+bc+ac}+\frac{b^3}{c^2+ab+bc+ac}+\frac{c^3}{a^2+ab+bc+ac}\)
\(=\frac{a^3}{(b+c)(b+a)}+\frac{b^3}{(c+a)(c+b)}+\frac{c^3}{(a+b)(a+c)}\)
Áp dụng BĐT AM-GM:
\(\frac{a^3}{(b+c)(b+a)}+\frac{b+c}{8}+\frac{b+a}{8}\geq 3\sqrt[3]{\frac{a^3}{64}}=\frac{3a}{4}\)
\(\frac{b^3}{(c+a)(c+b)}+\frac{c+a}{8}+\frac{c+b}{8}\geq 3\sqrt[3]{\frac{b^3}{64}}=\frac{3b}{4}\)
\(\frac{c^3}{(a+b)(a+c)}+\frac{a+b}{8}+\frac{a+c}{8}\geq 3\sqrt[3]{\frac{c^3}{64}}=\frac{3c}{4}\)
Cộng các BĐT trên vào và rút gọn:
\(\Rightarrow P+\frac{a+b+c}{2}\geq \frac{3}{4}(a+b+c)\)
\(\Rightarrow P\geq \frac{a+b+c}{4}(1)\)
Ta có một hệ quả quen thuộc của BĐT AM-GM đó là:
\((a+b+c)^2\geq 3(ab+bc+ac)\Leftrightarrow (a+b+c)^2\geq 9\)
\(\Rightarrow a+b+c\geq 3(2)\)
Từ \((1); (2)\Rightarrow P\geq \frac{3}{4}\) (đpcm)
Dấu bằng xảy ra khi \(a=b=c=1\)
Ta chứng minh \(P\ge-\dfrac{4}{3}\) hay
\(\dfrac{\left(a+b+c\right)^2}{30\left(a^2+b^2+c^2\right)}-\dfrac{1}{10}+\dfrac{a^3+b^3+c^3}{4abc}-\dfrac{3}{4}-\dfrac{131\left(a^2+b^2+c^2\right)}{60\left(ab+bc+ca\right)}+\dfrac{131}{60}\ge0\)
\(\Leftrightarrow\dfrac{\left(a+b+c\right)^2-3\left(a^2+b^2+c^2\right)}{30\left(a^2+b^2+c^2\right)}+\dfrac{a^3+b^3+c^3-3abc}{4abc}-\dfrac{131\left(a^2+b^2+c^2-ab-bc-ca\right)}{60\left(ab+bc+ca\right)}\ge0\)
\(\LeftrightarrowΣ_{cyc}\dfrac{-\left(a-b\right)^2}{30\left(a^2+b^2+c^2\right)}+Σ_{cyc}\dfrac{\dfrac{a+b+c}{2}\left(a-b\right)^2}{4abc}-Σ_{cyc}\dfrac{\dfrac{131}{2}\left(a-b\right)^2}{60\left(ab+bc+ca\right)}\ge0\)
\(\LeftrightarrowΣ_{cyc}\left(a-b\right)^2\left(\dfrac{\dfrac{a+b+c}{2}}{4abc}-\dfrac{\dfrac{131}{2}}{60\left(ab+bc+ca\right)}-\dfrac{1}{30\left(a^2+b^2+c^2\right)}\right)\ge0\)
\(ab+bc+ca=3abc\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=3\)
Đặt \(\dfrac{1}{a}=x;\dfrac{1}{b}=y;\dfrac{1}{c}=z\)\(\Rightarrow x+y+z=3\)
\(VT=\sum\dfrac{xyz}{yz+x^2}\le\sum\dfrac{xyz}{2x\sqrt{yz}}=\dfrac{1}{2}\sum\sqrt{yz}\le\dfrac{1}{2}\sum x=\dfrac{3}{2}\)
Áp dụng bđt AM-GM:
\(M\ge\dfrac{a^3}{a^2+\dfrac{a^2+b^2}{2}+b^2}+\dfrac{b^3}{b^2+\dfrac{b^2+c^2}{2}+c^2}+\dfrac{c^3}{c^2+\dfrac{a^2+c^2}{2}+a^2}\)
\(=\dfrac{a^3}{\dfrac{3}{2}\left(a^2+b^2\right)}+\dfrac{b^3}{\dfrac{3}{2}\left(b^2+c^2\right)}+\dfrac{c^3}{\dfrac{3}{2}\left(c^2+a^2\right)}\)
\(=\dfrac{2}{3}\left(\dfrac{a^3}{a^2+b^2}+\dfrac{b^3}{b^2+c^2}+\dfrac{c^3}{c^2+a^2}\right)\)
Xét:
\(\dfrac{a^3}{a^2+b^2}+\dfrac{b^3}{b^2+c^2}+\dfrac{c^3}{c^2+a^2}\)
\(=a-\dfrac{ab^2}{a^2+b^2}+b-\dfrac{b^2c}{b^2+c^2}+c-\dfrac{c^2a}{c^2+a^2}\)
\(\ge a+b+c-\dfrac{ab^2}{2ab}-\dfrac{b^2c}{2bc}-\dfrac{c^2a}{2ac}=a+b+c-\dfrac{a}{2}-\dfrac{b}{2}-\dfrac{c}{2}=\dfrac{a+b+c}{2}=\dfrac{3}{2}\)
\(\Leftrightarrow M\ge1."="\Leftrightarrow a=b=c=1\)
dòng thứ 5 từ dưới lên cái đầu là bc^2 nhé. Cái sau là ca^2