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a^2=bc
=>a*a=b/c
=>a/b=c/a=k
=>a=bk; c=ak
\(\dfrac{2022a+2021b}{2023a-2024b}=\dfrac{2022\cdot bk+2021b}{2023\cdot bk-2024b}=\dfrac{2022k+2021}{2023k-2024}\)
\(\dfrac{2022c+2021a}{2023c-2024a}=\dfrac{2022ak+2021a}{2023ak-2024a}=\dfrac{2022k+2021}{2023k-2024}\)
=>\(\dfrac{2022a+2021b}{2023a-2024b}=\dfrac{2022c+2021a}{2023c-2024a}\)
a: \(\left|a-2b+3\right|^{2023}>=0\forall a,b\)
\(\left(b-1\right)^{2024}>=0\forall b\)
Do đó: \(\left|a-2b+3\right|^{2023}+\left(b-1\right)^{2024}>=0\forall a,b\)
Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}a-2b+3=0\\b-1=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}b=1\\a=2b-3=2\cdot1-3=-1\end{matrix}\right.\)
Thay a=-1 và b=1 vào P, ta được:
\(P=\left(-1\right)^{2023}\cdot1^{2024}+2024=2024-1=2023\)
\(\dfrac{ab}{a+3b+2c}=\dfrac{ab}{\left(a+c\right)+\left(b+c\right)+2b}\le\dfrac{1}{9}\left(\dfrac{ab}{a+c}+\dfrac{ab}{b+c}+\dfrac{ab}{2b}\right)\)
\(=\dfrac{1}{9}\left(\dfrac{ab}{a+c}+\dfrac{ab}{b+c}+\dfrac{a}{2}\right)\)
Tương tự:
\(\dfrac{bc}{b+3c+2a}\le\dfrac{1}{9}\left(\dfrac{bc}{a+b}+\dfrac{bc}{a+c}+\dfrac{b}{2}\right)\)
\(\dfrac{ac}{c+3a+2b}\le\dfrac{1}{9}\left(\dfrac{ac}{b+c}+\dfrac{ac}{a+b}+\dfrac{c}{2}\right)\)
Cộng vế:
\(P\le\dfrac{1}{9}\left(\dfrac{bc+ac}{a+b}+\dfrac{bc+ab}{a+c}+\dfrac{ab+ac}{b+c}+\dfrac{a+b+c}{2}\right)\)
\(P\le\dfrac{1}{9}.\left(a+b+c+\dfrac{a+b+c}{2}\right)=\dfrac{1}{2}\)
Dấu "=" xảy ra khi \(a=b=c=1\)
a+b−cc=b+c−aa=c+a−bb
⇒a+b−cc+1=b+c−aa+1=c+a−bb+1
⇒a+bc=b+ca=c+ab
+)Nếu a+b+c=0⇒a+b=−c;b+c=−a;c+a=−b
⇒B=a+ba.c+ac.b+cb=−ca.−bc.−ab=−(abc)abc=−1
Nếu a+b+c≠0
Áp dụng tính chất dãy tỉ số bằng nhau ta có
a+bc=b+ca=c+ab=2(a+b+c)a+b+c=2
⇒a+b=2c
b+c=2a
c+a=2b
⇒B=2ca.2bc.2ab=2.2.2=8
\(2ab+6bc+2ac=7abc\Rightarrow\dfrac{6}{a}+\dfrac{2}{b}+\dfrac{2}{c}=7\)
Đặt \(\left(\dfrac{2}{a};\dfrac{1}{b};\dfrac{1}{c}\right)=\left(x;y;z\right)\Rightarrow3x+2y+2z=7\)
\(C=\dfrac{4}{\dfrac{2}{a}+\dfrac{1}{b}}+\dfrac{9}{\dfrac{4}{a}+\dfrac{1}{c}}+\dfrac{4}{\dfrac{1}{b}+\dfrac{1}{c}}=\dfrac{4}{x+y}+\dfrac{9}{2x+z}+\dfrac{4}{y+z}\)
\(C\ge\dfrac{\left(2+3+2\right)^2}{x+y+2x+z+y+z}=\dfrac{49}{7}=7\)
Dấu "=" xảy ra khi \(x=y=z=1\) hay \(\left(a;b;c\right)=\left(2;1;1\right)\)
Lời giải:
$(\frac{2023a}{2024c})^3=(\frac{2024b}{2025a})^3=(\frac{2025c}{2023b})^3=\frac{2023a}{2024b}.\frac{2024b}{2025a}.\frac{2025c}{2023b}=1$
$\Rightarrow \frac{2023a}{2024c}=\frac{2024b}{2025a}=\frac{2025c}{2023b}=1$
$\Rightarrow 2023a=2024c; 2024b=2025a; 2025c=2023b$
Do đó:
$\frac{2023a}{506c}+\frac{2024b}{675a}+\frac{2025c}{289b}=\frac{2024c}{506c}+\frac{2025a}{675a}+\frac{2023b}{289b}$
$=4+3+7=14$