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\(VT=\dfrac{a^2}{b+ab^2c}+\dfrac{b^2}{b+abc^2}+\dfrac{c^2}{c+a^2bc}\ge\dfrac{\left(a+b+c\right)^2}{a+b+c+abc\left(a+b+c\right)}=\dfrac{9}{3+3abc}\)
\(VT\ge\dfrac{9}{3+\dfrac{\left(a+b+c\right)^3}{9}}=\dfrac{3}{2}\)
Dấu "=" xảy ra khi \(a=b=c=1\)
Áp dụng bất đẳng thức \(\dfrac{9}{x+y+z}\le\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\) với x, y, z > 0 ta có:
\(\dfrac{1}{2a+b}+\dfrac{1}{2b+c}+\dfrac{1}{2c+a}=\dfrac{1}{9}\left(\dfrac{9}{a+a+b}+\dfrac{9}{b+b+c}+\dfrac{1}{c+c+a}\right)\le\dfrac{1}{9}\left(\dfrac{1}{a}+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{b}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{c}+\dfrac{1}{c}+\dfrac{1}{a}\right)=\dfrac{1}{9}.3\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=\dfrac{1}{3a}+\dfrac{1}{3b}+\dfrac{1}{3c}\).
Ta có:
\(\dfrac{2a+b}{a+b}+\dfrac{2c+d}{c+d}+\dfrac{2b+c}{b+c}+\dfrac{2d+a}{d+a}=6\)
⇔ \(\left(\dfrac{2a+b}{a+b}-1\right)+\left(\dfrac{2c+d}{c+d}-1\right)+\left(\dfrac{2b+c}{b+c}-1\right)+\left(\dfrac{2d+a}{d+a}-1\right)=2\)
⇔ \(\dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+d}+\dfrac{d}{d+a}=2\)
⇔ \(\left(1-\dfrac{a}{a+b}\right)-\dfrac{b}{b+c}+\left(1-\dfrac{c}{c+d}\right)-\dfrac{d}{d+a}=0\)
⇔ \(\dfrac{b}{a+b}-\dfrac{b}{b+c}+\dfrac{d}{c+d}-\dfrac{d}{d+a}=0\)
⇔ \(\dfrac{b\left(b+c\right)-b\left(a+b\right)}{\left(a+b\right)\left(b+c\right)}+\dfrac{d\left(d+a\right)-d\left(c+d\right)}{\left(c+d\right)\left(d+a\right)}=0\)
⇔ \(\dfrac{b\left(c-a\right)}{\left(a+b\right)\left(b+c\right)}+\dfrac{d\left(a-c\right)}{\left(c+d\right)\left(d+a\right)}=0\)
⇔ \(\dfrac{b\left(c-a\right)}{\left(a+b\right)\left(b+c\right)}-\dfrac{d\left(c-a\right)}{\left(c+d\right)\left(d+a\right)}=0\)
⇔ \(\left(c-a\right)\left(\dfrac{b}{\left(a+b\right)\left(b+c\right)}-\dfrac{d}{\left(c+d\right)\left(d+a\right)}\right)=0\)
⇒ \(\dfrac{b}{\left(a+b\right)\left(b+c\right)}-\dfrac{d}{\left(c+d\right)\left(d+a\right)}=0\) \(\left(a\ne c\right)\)
⇒ \(b\left(c+d\right)\left(d+a\right)-d\left(a+b\right)\left(b+c\right)=0\)
⇔ \(\left(bc+bd\right)\left(d+a\right)-\left(ad+bd\right)\left(b+c\right)=0\)
⇔ \(bcd+abc+bd^2+abd-abd-acd-b^2d-bcd=0\)
⇔ \(abc+bd^2-acd-b^2d=0\)
⇔ \(ac\left(b-d\right)-bd\left(b-d\right)=0\)
⇔ \(\left(b-d\right)\left(ac-bd\right)=0\)
⇒ \(ac-bd=0\) \(\left(b\ne d\right)\)
⇔ \(ac=bd\)
Khi đó:
\(A=abcd=\left(ac\right)^2\)
⇒ \(ĐPCM\)
Hi vọng là tìm GTLN:
Không mất tính tổng quát, giả sử b, c cùng phía với 1 \(\Rightarrow\left(b-1\right)\left(c-1\right)\ge0\Leftrightarrow bc\ge b+c-1\).
Áp dụng bất đẳng thức AM - GM ta có:
\(4=a^2+b^2+c^2+abc\ge a^2+2bc+abc\Leftrightarrow2bc+abc\le4-a^2\Leftrightarrow bc\left(a+2\right)\le\left(2-a\right)\left(a+2\right)\Leftrightarrow bc+a\le2\)
\(\Rightarrow a+b+c\le3\).
Áp dụng bất đẳng thức Schwarz ta có:
\(P\le\dfrac{ab}{9}\left(\dfrac{1}{a}+\dfrac{2}{b}\right)+\dfrac{bc}{9}\left(\dfrac{1}{b}+\dfrac{2}{c}\right)+\dfrac{ca}{9}\left(\dfrac{1}{c}+\dfrac{2}{a}\right)=\dfrac{1}{9}.3\left(a+b+c\right)=\dfrac{1}{3}\left(a+b+c\right)\le1\).
Đẳng thức xảy ra khi a = b = c = 1.
Điều kiện đã cho có thể được viết lại thành \(\dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+d}+\dfrac{d}{d+a}=2\)
hay \(1-\dfrac{a}{a+b}-\dfrac{b}{b+c}+1-\dfrac{c}{c+d}-\dfrac{d}{d+a}=0\)
\(\Leftrightarrow\dfrac{b}{a+b}-\dfrac{b}{b+c}+\dfrac{d}{c+d}-\dfrac{d}{d+a}=0\)
\(\Leftrightarrow\dfrac{b^2+bc-ab-b^2}{\left(a+b\right)\left(b+c\right)}+\dfrac{d^2+da-cd-d^2}{\left(c+d\right)\left(d+a\right)}=0\)
\(\Leftrightarrow\dfrac{b\left(c-a\right)}{\left(a+b\right)\left(b+c\right)}+\dfrac{d\left(a-c\right)}{\left(c+d\right)\left(d+a\right)}=0\)
\(\Leftrightarrow\left(c-a\right)\left[\dfrac{b}{\left(a+b\right)\left(b+c\right)}-\dfrac{d}{\left(c+d\right)\left(d+a\right)}\right]=0\)
\(\Leftrightarrow\dfrac{b}{\left(a+b\right)\left(b+c\right)}=\dfrac{d}{\left(c+d\right)\left(d+a\right)}\) (do \(c\ne a\))
\(\Leftrightarrow b\left(cd+ca+d^2+da\right)=d\left(ab+ac+b^2+bc\right)\)
\(\Leftrightarrow bcd+abc+bd^2+abd=abd+acd+b^2d+bcd\)
\(\Leftrightarrow abc+bd^2-acd-b^2d=0\)
\(\Leftrightarrow ac\left(b-d\right)-bd\left(b-d\right)=0\)
\(\Leftrightarrow\left(b-d\right)\left(ac-bd\right)=0\)
\(\Leftrightarrow ac=bd\) (do \(b\ne d\))
Do đó \(A=abcd=ac.ac=\left(ac\right)^2\), mà \(a,c\inℕ^∗\) nên A là SCP (đpcm)
Áp dụng BĐT Cauchy cho 3 số dương a , b , c , ta có :
\(D=\dfrac{a}{a+2b}+\dfrac{b}{b+2c}+\dfrac{c}{c+2a}=\dfrac{a^2}{a^2+2ab}+\dfrac{b^2}{b^2+2bc}+\dfrac{c^2}{c^2+2ac}\ge\dfrac{\left(a+b+c\right)^2}{a^2+b^2+c^2+2ab+2bc+2ac}=\dfrac{\left(a+b+c\right)^2}{\left(a+b+c\right)^2}=1\)Dấu " = " xảy ra \(\Leftrightarrow a=b=c=1\)