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Bạn CM \(a^5+b^5\ge ab\left(a^3+b^3\right)\)
\(\Rightarrow\frac{ab}{a^5+b^5+ab}\le\frac{1}{a^3+b^3+abc}\)
Tiếp tục \(a^3+b^3\ge ab\left(a+b\right)\)
\(\Rightarrow\frac{1}{a^3+b^3+abc}\le\frac{1}{ab\left(a+b\right)+abc}=\frac{c}{a+b+c}\)
\(\Rightarrow\frac{ab}{a^5+b^5+ab}\le\frac{c}{a+b+c}\)
Tương tự cộng lại suy ra \(VT\le1\)
Dấu = xảy ra khi a=b=c=1
Thỏa mãn $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1$ hay $a+b+c=1$ vậy bạn?
- Theo BĐT Cauchy ta có:
\(\sqrt{a.1}\le\dfrac{a+1}{2}\)
\(\sqrt{b.1}\le\dfrac{b+1}{2}\)
\(\sqrt{c.1}\le\dfrac{c+1}{2}\)
\(\sqrt{ab}\le\dfrac{a+b}{2}\)
\(\sqrt{bc}\le\dfrac{b+c}{2}\)
\(\sqrt{ca}\le\dfrac{c+a}{2}\)
\(\Rightarrow\sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\le\dfrac{3\left(a+b+c\right)+3}{2}=\dfrac{3.3+3}{2}=6\)
Dấu "=" xảy ra khi \(a=b=c=1\)
Mà ta có: \(\sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{ab}+\sqrt{bc}+\sqrt{ca}=6\)
\(\Rightarrow a=b=c=1\)
\(M=\dfrac{a^{30}+b^4+c^{1975}}{a^{30}+b^4+c^{2023}}=\dfrac{1^{30}+1^4+1^{1975}}{1^{30}+1^4+1^{2023}}=1\)
\(A=\frac{ab}{a+c+b+c}+\frac{bc}{a+b+a+c}+\frac{ca}{a+b+b+c}\)
\(\le\frac{1}{4}\left(\frac{ab}{a+c}+\frac{ab}{b+c}+\frac{bc}{a+b}+\frac{bc}{a+c}+\frac{ca}{a+b}+\frac{ca}{b+c}\right)\)
\(=\frac{1}{4}\left(a+b+c\right)=\frac{1}{4}\)
Nên max A là \(\frac{1}{4}\) khi \(a=b=c=\frac{1}{3}\)
\(a^2+b^2-ab\ge\dfrac{1}{2}\left(a+b\right)^2-\dfrac{1}{4}\left(a+b\right)^2=\dfrac{1}{4}\left(a+b\right)^2\)
\(\Rightarrow\dfrac{1}{\sqrt{a^2-ab+b^2}}\le\dfrac{1}{\sqrt{\dfrac{1}{4}\left(a+b\right)^2}}=\dfrac{2}{a+b}\le\dfrac{1}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\)
Tương tự:
\(\dfrac{1}{\sqrt{b^2-bc+c^2}}\le\dfrac{1}{2}\left(\dfrac{1}{b}+\dfrac{1}{c}\right)\) ; \(\dfrac{1}{\sqrt{c^2-ca+a^2}}\le\dfrac{1}{2}\left(\dfrac{1}{c}+\dfrac{1}{a}\right)\)
Cộng vế:
\(P\le\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=3\)
Dấu "=" xảy ra khi \(a=b=c=1\)
Lời giải:
\(Q=\frac{ab}{c+ab}+\frac{ac}{b+ac}+\frac{bc}{a+bc}-\frac{1}{4abc}=\frac{ab}{c(a+b+c)+ab}+\frac{ac}{b(a+b+c)+ac}+\frac{bc}{a(a+b+c)+bc}-\frac{1}{4abc}\)
\(=\frac{ab}{(c+a)(c+b)}+\frac{ac}{(b+a)(b+c)}+\frac{bc}{(a+b)(a+c)}-\frac{1}{4abc}\)
\(=\frac{ab(a+b)+ac(a+c)+bc(b+c)}{(a+b)(b+c)(c+a)}-\frac{1}{4abc}\)
\(=\frac{(a+b)(b+c)(c+a)-2abc}{(a+b)(b+c)(c+a)}-\frac{1}{4abc}\) (đẳng thức quen thuộc \((a+b)(b+c)(c+a)=ab(a+b)+bc(b+c)+ca(c+a)+2abc\) )
\(=1-\left(\frac{2abc}{(a+b)(b+c)(c+a)}+\frac{1}{4abc}\right)\)
Áp dụng BĐT AM-GM:
\(\frac{2abc}{(a+b)(b+c)(c+a)}+\frac{1}{108abc}\geq 2\sqrt{\frac{1}{54(a+b)(b+c)(c+a)}}\).
Mà \(2=(a+b)+(b+c)+(c+a)\geq 3\sqrt[3]{(a+b)(b+c)(c+a)}\Rightarrow (a+b)(b+c)(c+a)\leq \frac{8}{27}\)
\(\Rightarrow \frac{2abc}{(a+b)(b+c)(c+a)}+\frac{1}{108abc}\geq \frac{1}{2}\)
\(1=a+b+c\geq 3\sqrt[3]{abc}\Rightarrow abc\leq \frac{1}{27}\)
\(\Rightarrow \frac{13}{54abc}\geq \frac{13}{2}\)
Do đó: \(\frac{2abc}{(a+b)(b+c)(c+a)}+\frac{1}{4abc}\geq 7\)
\(\Rightarrow Q\leq 1-7=-6=Q_{\max}\)
Dấu "=" xảy ra khi $a=b=c=\frac{1}{3}$
bạn ơi lí do vì sao ở cái biểu thức bạn rút gọn là \(1-\left(\dfrac{2abc}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}+\dfrac{1}{4abc}\right)\)
nhưng bạn dùng bđt cô-si lại là
\(\dfrac{2abc}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}+\dfrac{1}{108abc}\)
\(\dfrac{1}{4abc}\) bạn không dùng mà bạn lại dùng là \(\dfrac{1}{108abc}\) vậy bạn?
Bạn có thể giải thích rõ chỗ đó cho mình được không bạn?
\(Q=ac+bc-2022ab\le ac+bc=c\left(a+b\right)\le\dfrac{1}{4}\left(c+a+b\right)^2=\dfrac{1}{4}\)
\(Q_{max}=\dfrac{1}{4}\) khi \(\left\{{}\begin{matrix}a+b+c=1\\ab=0\\c=a+b\end{matrix}\right.\) \(\Leftrightarrow\left(a;b;c\right)=\left(0;\dfrac{1}{2};\dfrac{1}{2}\right);\left(\dfrac{1}{2};0;\dfrac{1}{2}\right)\)
\(Q=c\left(a+b\right)-2022ab\ge c\left(a+b\right)-\dfrac{1011}{2}\left(a+b\right)^2\)
\(Q\ge c\left(1-c\right)-\dfrac{1011}{2}\left(1-c\right)^2\)
\(Q\ge c\left(1-c\right)-\dfrac{1011}{2}c\left(c-2\right)-\dfrac{1011}{2}\)
\(Q\ge\dfrac{c\left(1011+1013\left(1-c\right)\right)}{2}-\dfrac{1011}{2}\ge-\dfrac{1011}{2}\)
\(Q_{min}=-\dfrac{1011}{2}\) khi \(\left(a;b;c\right)=\left(\dfrac{1}{2};\dfrac{1}{2};0\right)\)
Từ giả thiết:
\(2024abc\ge a^2+b^2+c^2\ge3\sqrt[3]{a^2b^2c^2}\Rightarrow abc\ge\dfrac{3^3}{2024^3}\)
Lại có:
\(2024abc\ge a^2+b^2+c^2\ge\dfrac{1}{3}\left(a+b+c\right)^2\ge\dfrac{1}{3}\left(a+b+c\right).3\sqrt[3]{abc}\ge a+b+c.\sqrt[3]{\dfrac{3^3}{2024^3}}\)
\(\Rightarrow2024abc\ge\dfrac{3}{2024}\left(a+b+c\right)\)
\(\Rightarrow\dfrac{a+b+c}{abc}\le\dfrac{2024^2}{3}\)
Từ đó:
\(Q=\dfrac{a}{a^2+bc}+\dfrac{b}{b^2+ca}+\dfrac{c}{c^2+ab}\)
\(Q\le\dfrac{a}{2\sqrt{a^2.bc}}+\dfrac{b}{2\sqrt{b^2.ca}}+\dfrac{c}{2\sqrt{c^2.ab}}=\dfrac{1}{2}\left(\dfrac{1}{\sqrt{bc}}+\dfrac{1}{\sqrt{ca}}+\dfrac{1}{\sqrt{ab}}\right)\)
\(Q\le\dfrac{1}{2}\left(\dfrac{\sqrt{a}+\sqrt{b}+\sqrt{c}}{\sqrt{abc}}\right)\le\dfrac{\sqrt{3\left(a+b+c\right)}}{2\sqrt{abc}}=\dfrac{\sqrt{3}}{2}.\sqrt{\dfrac{a+b+c}{abc}}\le\dfrac{\sqrt{3}}{2}.\sqrt{\dfrac{2024^2}{3}}=1012\)
\(Q_{max}=1012\) khi \(a=b=c=\dfrac{3}{2024}\)