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Lời giải:
Áp dụng BĐT AM-GM ta có:
\(2a+b+c=(a+b)+(a+c)\geq 2\sqrt{(a+b)(a+c)}\)
\(\Rightarrow (2a+b+c)^2\geq 4(a+b)(a+c)\)
\(\Rightarrow \frac{1}{(2a+b+c)^2}\leq \frac{1}{4(a+b)(a+c)}\)
Hoàn toàn tương tự với các phân thức còn lại suy ra:
\(P\leq \frac{1}{4}\left(\frac{1}{(a+b)(a+c)}+\frac{1}{(b+c)(b+a)}+\frac{1}{(c+a)(c+b)}\right)\)
\(\Leftrightarrow P\leq \frac{1}{4}.\frac{(b+c)+(c+a)+(a+b)}{(a+b)(b+c)(c+a)}\)
\(\Leftrightarrow P\leq \frac{a+b+c}{2(a+b)(b+c)(c+a)}\)
Lại có: \((a+b)(b+c)(c+a)\geq 2\sqrt{ab}.2\sqrt{bc}.2\sqrt{ac}=8abc\) (theo AM-GM)
\(\Rightarrow P\leq \frac{a+b+c}{2.8abc}=\frac{a+b+c}{16abc}(1)\)
Tiếp tục áp dụng BĐT AM-GM:
\(\frac{1}{a^2}+\frac{1}{b^2}\geq \frac{2}{ab}; \frac{1}{b^2}+\frac{1}{c^2}\geq \frac{2}{bc}; \frac{1}{c^2}+\frac{1}{a^2}\geq \frac{2}{ac}\)
\(\Rightarrow 2\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\geq 2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right)\)
\(\Leftrightarrow 3\geq \frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=\frac{a+b+c}{abc}\)
\(\Rightarrow a+b+c\leq 3abc(2)\)
Từ \((1); (2)\Rightarrow P\leq \frac{3abc}{16abc}=\frac{3}{16}\)
Vậy \(P_{\max}=\frac{3}{16}\). Dấu bằng xảy ra khi \(a=b=c=1\)
bai nay t lam roi vao trang chu cua nick thangbnsh cua t keo xuong tim la thay
Câu hỏi của Tuyển Trần Thị - Toán lớp 9 | Học trực tuyến
Lời giải:
Áp dụng BĐT Cauchy-Schwarz:
\(P=\frac{1}{a^3(b+c)}+\frac{1}{b^3(a+c)}+\frac{1}{c^3(a+b)}\)
\(=\frac{\left(\frac{1}{a}\right)^2}{a(b+c)}+\frac{\left(\frac{1}{b}\right)^2}{b(a+c)}+\frac{\left(\frac{1}{c}\right)^2}{c(a+b)}\)
\(\geq \frac{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2}{a(b+c)+b(a+c)+c(a+b)}=\frac{(ab+bc+ac)^2}{2(ab+bc+ac)}=\frac{ab+bc+ac}{2}\) (thay $1=abc$)
Mà theo BĐT AM-GM:
\(ab+bc+ac\geq 3\sqrt[3]{(abc)^2}=3\). Do đó:
\(P\geq \frac{ab+bc+ac}{2}\geq \frac{3}{2}\)
Vậy \(P_{\min}=\frac{3}{2}\Leftrightarrow a=b=c=1\)
Cách khác:
Áp dụng BĐT AM-GM:
\(\frac{1}{a^3(b+c)}+\frac{a(b+c)}{4}\geq 2\sqrt{\frac{1}{a^3(b+c)}.\frac{a(b+c)}{4}}=\frac{1}{a}=\frac{abc}{a}=bc\)
Tương tự:
\(\frac{1}{b^3(a+c)}+\frac{b(a+c)}{4}\geq ac\)
\(\frac{1}{c^3(a+b)}+\frac{c(a+b)}{4}\geq ab\)
Cộng theo vế các BĐT trên ta có:
\(P+\frac{ab+bc+ac}{2}\geq ab+bc+ac\)
\(\Rightarrow P\geq \frac{ab+bc+ac}{2}\geq \frac{3\sqrt[3]{(abc)^2}}{2}=\frac{3}{2}\)
Vậy \(P_{\min}=\frac{3}{2}\Leftrightarrow a=b=c=1\)
\(\dfrac{1}{\left(a+b+a+c\right)^2}\le\dfrac{1}{4\left(a+b\right)\left(a+c\right)}=\dfrac{1}{4\left(a^2+ab+bc+ca\right)}\le\dfrac{1}{64}\left(\dfrac{1}{a^2}+\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\right)\)
\(\le\dfrac{1}{64}\left(\dfrac{1}{a^2}+\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)=\dfrac{1}{64}\left(\dfrac{2}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)\)
Tương tự và cộng lại:
\(P\le\dfrac{1}{64}\left(\dfrac{4}{a^2}+\dfrac{4}{b^2}+\dfrac{4}{c^2}\right)=\dfrac{1}{16}.3=\dfrac{3}{16}\)
Dấu "=" xảy ra khi \(a=b=c=1\)
Áp dụng bđt: \(\dfrac{1}{x+y}\le\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\left(1\right)\)
\(\dfrac{1}{2a+b+c}=\dfrac{1}{\left(a+b\right)+\left(a+c\right)}\le\dfrac{1}{4}\left(\dfrac{1}{a+b}+\dfrac{1}{a+c}\right)\)
\(\Rightarrow P\le\dfrac{1}{16}\left[\left(\dfrac{1}{a+b}+\dfrac{1}{a+c}\right)^2+\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}\right)^2+\left(\dfrac{1}{b+c}+\dfrac{1}{a+c}\right)^2\right]\)\(\Rightarrow16P\le\dfrac{2}{\left(a+b\right)^2}+\dfrac{2}{\left(b+c\right)^2}+\dfrac{2}{\left(a+c\right)^2}+\dfrac{2}{\left(a+b\right)\left(b+c\right)}+\dfrac{2}{\left(a+b\right)\left(b+c\right)}+\dfrac{2}{\left(b+c\right)\left(c+a\right)}\)
Áp dụng: \(x^2+y^2+z^2\ge xy+yz+xz\left(2\right)\) với a+b=x,b+c=y,c+a=z
\(\Rightarrow16P\le\dfrac{4}{\left(a+b\right)^2}+\dfrac{4}{\left(b+c\right)^2}+\dfrac{4}{\left(c+a\right)^2}\)
Ta có: \(\dfrac{1}{\left(a+b\right)^2}\le4.16.\left(\dfrac{1}{a}+\dfrac{1}{b}\right)^2\)(do (1))
\(\Rightarrow16P\le\dfrac{1}{4}.16\left[\left(\dfrac{1}{a}+\dfrac{1}{b}\right)^2+\left(\dfrac{1}{b}+\dfrac{1}{c}\right)^2+\left(\dfrac{1}{c}+\dfrac{1}{a}\right)^2\right]=\dfrac{1}{4}\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+\dfrac{2}{ab}+\dfrac{2}{bc}+\dfrac{2}{ca}\right)\le\dfrac{1}{4}.4.\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)=3\)(do(2) và \(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}=3\))
\(\Rightarrow P\le\dfrac{3}{16}\)
\(ĐTXR\Leftrightarrow a=b=c=1\)
Áp dụng BĐT AM-GM ta có:
\(\dfrac{a^3}{\left(1+b\right)\left(1+c\right)}+\dfrac{b+1}{8}+\dfrac{c+1}{8}\)
\(\ge3\sqrt[3]{\dfrac{a^3}{\left(1+b\right)\left(1+c\right)}\cdot\dfrac{b+1}{8}\cdot\dfrac{c+1}{8}}=\dfrac{3a}{4}\)
Tương tự cho 2 BĐT còn lại ta cũng có:
\(\dfrac{b^3}{\left(1+c\right)\left(1+a\right)}+\dfrac{c+1}{8}+\dfrac{a+1}{8}\ge\dfrac{3b}{4};\dfrac{c^3}{\left(1+a\right)\left(1+b\right)}+\dfrac{a+1}{8}+\dfrac{b+1}{8}\ge\dfrac{3c}{4}\)
Cộng theo vế 3 BĐT trên ta có:
\(VT+\dfrac{2\left(a+b+c+3\right)}{8}\ge\dfrac{3\left(a+b+c\right)}{4}\)
\(\Leftrightarrow VT+\dfrac{2\left(3\sqrt[3]{abc}+3\right)}{8}\ge\dfrac{3\cdot3\sqrt[3]{abc}}{4}\Leftrightarrow VT\ge\dfrac{3}{4}=VP\)
Khi \(a=b=c=1\)
Lời giải:
Áp dụng BĐT AM-GM ta có:
\(\frac{a^3}{(b+2)(c+3)}+\frac{b+2}{36}+\frac{c+3}{48}\geq 3\sqrt[3]{\frac{a^3}{36.48}}=\frac{a}{4}\)
Tương tự:\(\frac{b^3}{(c+2)(a+3)}+\frac{c+2}{36}+\frac{a+3}{48}\geq \frac{b}{4}\)
\(\frac{c^3}{(a+2)(b+3)}+\frac{a+2}{36}+\frac{b+3}{48}\geq \frac{c}{4}\)
Cộng theo vế các BĐT trên và rút gọn ta có:
\(\frac{a^3}{(b+2)(c+3)}+\frac{b^3}{(c+2)(a+3)}+\frac{c^3}{(a+2)(b+3)}\geq \frac{29}{144}(a+b+c)-\frac{17}{48}\)
Mà cũng theo AM-GM:
\(a+b+c\geq 3\sqrt[3]{abc}=3\)
\(\Rightarrow \frac{a^3}{(b+2)(c+3)}+\frac{b^3}{(c+2)(a+3)}+\frac{c^3}{(a+2)(b+3)}\geq \frac{29}{144}(a+b+c)-\frac{17}{48}\geq \frac{29}{144}.3-\frac{17}{48}=\frac{1}{4}\)
Ta có đpcm
Dấu "=" xảy ra khi $a=b=c=1$
Áp dụng BĐT AM-GM ta có:
\(\dfrac{3}{2}\ge a+b+c\ge3\sqrt[3]{abc}\Rightarrow\dfrac{1}{2}\ge\sqrt[3]{abc}\Rightarrow\dfrac{1}{8}\ge abc\)
Áp dụng BĐT Holder ta có:
\(B=\left(3+\dfrac{1}{a}+\dfrac{1}{b}\right)\left(3+\dfrac{1}{b}+\dfrac{1}{c}\right)\left(3+\dfrac{1}{c}+\dfrac{1}{a}\right)\)
\(\ge\left(\sqrt[3]{3\cdot3\cdot3}+\sqrt[3]{\dfrac{1}{a}\cdot\dfrac{1}{b}\cdot\dfrac{1}{c}}+\sqrt[3]{\dfrac{1}{a}\cdot\dfrac{1}{b}\cdot\dfrac{1}{c}}\right)^3\)
\(=\left(3+2\sqrt[3]{\dfrac{1}{abc}}\right)^3\ge\left(3+2\sqrt[3]{\dfrac{1}{\dfrac{1}{8}}}\right)^3=343\)
Khi \(a=b=c=\dfrac{1}{2}\)