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\(P=\dfrac{a^3}{a^2+ab+b^2}+\dfrac{b^3}{b^2+bc+c^2}+\dfrac{c^3}{c^2+ca+a^2}\le\dfrac{2}{3}\left(\dfrac{a^3}{a^2+b^2}+\dfrac{b^3}{b^2+c^2}+\dfrac{c^3}{c^2+a^2}\right)\le\dfrac{2}{3}\left[\left(a+b+c\right)-\dfrac{a+b+c}{2}\right]=\dfrac{2}{3}\left(2019-\dfrac{2019}{2}\right)=673\)
Giải:
Ta có:
\(P=\dfrac{a^3}{\sqrt{1+b^2}}+\dfrac{b^3}{\sqrt{1+c^2}}+\dfrac{c^3}{\sqrt{1+a^2}}\)
\(\Leftrightarrow P+3=\dfrac{a^3}{\sqrt{1+b^2}}+b^2+\dfrac{b^3}{\sqrt{1+c^2}}+c^2\dfrac{c^3}{\sqrt{1+a^2}}+a^2\)
\(\Leftrightarrow P+\dfrac{6}{4\sqrt{2}}=\dfrac{a^3}{2\sqrt{1+b^2}}+\dfrac{a^2}{2\sqrt{1+b^2}}+\dfrac{1+b^2}{4\sqrt{2}}+\dfrac{b^3}{2\sqrt{1+c^2}}+\dfrac{b^2}{2\sqrt{1+c^2}}+\dfrac{1+c^2}{4\sqrt{2}}+\dfrac{c^3}{2\sqrt{1+a^2}}+\dfrac{c^2}{2\sqrt{1+a^2}}+\dfrac{1+a^2}{4\sqrt{2}}\)
\(\ge3\sqrt[3]{\dfrac{a^6}{16\sqrt{2}}}+3\sqrt[3]{\dfrac{b^6}{16\sqrt{2}}}+3\sqrt[3]{\dfrac{c^6}{16\sqrt{2}}}\)
\(\Rightarrow P+\dfrac{3}{2\sqrt{2}}\ge\dfrac{3}{2\sqrt[3]{2\sqrt{2}}}\left(a^2+b^2+c^2\right)=\dfrac{9}{2\sqrt[6]{8}}\)
\(\Rightarrow P\ge\dfrac{9}{2\sqrt[6]{2^3}}-\dfrac{3}{2\sqrt{2}}=\dfrac{9}{2\sqrt{2}}-\dfrac{3}{2\sqrt{2}}=\dfrac{3}{\sqrt{2}}\)
Dấu "=" xảy ra khi \(a=b=c=1\)
Áp dụng BĐt cauchy-schwarz:(dạng phân thức + đa thức )
\(P=\dfrac{a^3}{\sqrt{1+b^2}}+\dfrac{b^3}{\sqrt{1+c^2}}+\dfrac{c^3}{\sqrt{1+a^2}}=\dfrac{a^4}{a\sqrt{1+b^2}}+\dfrac{b^4}{b\sqrt{1+c^2}}+\dfrac{c^4}{c\sqrt{1+a^2}}\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{a\sqrt{1+b^2}+b\sqrt{1+c^2}+c\sqrt{1+a^2}}\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{\sqrt{\left(a^2+b^2+c^2\right)\left(3+a^2+b^2+c^2\right)}}=\dfrac{9}{\sqrt{18}}=\dfrac{3}{\sqrt{2}}=\dfrac{3\sqrt{2}}{2}\)
dấu = xảy ra khi a=b=c=1
Bài 1:ta có BĐt \(a^3+b^3\ge ab\left(a+b\right)\)vì nó tương đương với \(\left(a+b\right)\left(a-b\right)^2\ge0\)(luôn đúng với a,b>0)
Áp dụng vào bài toán:
\(\dfrac{a^3+b^3}{2ab}+\dfrac{b^3+c^3}{2bc}+\dfrac{c^3+a^3}{2ac}\ge\dfrac{ab\left(a+b\right)}{2ab}+\dfrac{bc\left(b+c\right)}{2bc}+\dfrac{ca\left(c+a\right)}{2ac}=a+b+c\)dấu = xảy ra khi a=b=c
bài 2:
cần chứng minh \(\dfrac{a-b}{b+c}+\dfrac{b-c}{c+d}+\dfrac{c-d}{d+a}+\dfrac{d-a}{a+b}\ge0\)
hay \(\dfrac{a-b}{b+c}+1+\dfrac{b-c}{c+d}+1+\dfrac{c-d}{d+a}+1+\dfrac{d-a}{a+b}+1\ge4\)
\(\Leftrightarrow\dfrac{a+c}{b+c}+\dfrac{b+d}{c+d}+\dfrac{c+a}{d+a}+\dfrac{d+b}{a+b}\ge4\)
xét \(VT=\left(a+c\right)\left(\dfrac{1}{b+c}+\dfrac{1}{a+d}\right)+\left(b+d\right)\left(\dfrac{1}{c+d}+\dfrac{1}{a+b}\right)\)
Áp dụng BĐT cauchy dạng phân thức:
\(\dfrac{1}{b+c}+\dfrac{1}{a+d}\ge\dfrac{4}{a+b+c+d};\dfrac{1}{c+d}+\dfrac{1}{a+b}\ge\dfrac{4}{a+b+c+d}\)
do đó \(VT\ge\dfrac{4\left(a+c\right)}{a+b+c+d}+\dfrac{4\left(b+d\right)}{a+b+c+d}=4\)
dấu = xảy ra khi a=b=c=d
\(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{ab}{cd}\)
\(\Leftrightarrow\left(a^2+b^2\right)cd=ab\left(c^2+d^2\right)\)
\(\Leftrightarrow a^2cd-b^2cd=abc^2+abd^2\)
\(\Leftrightarrow a^2cd-abc^2-abd^2+b^2cd=0\)
\(\Leftrightarrow ac\left(ad-bc\right)-bd\left(ad-bc\right)=0\)
\(\Leftrightarrow\left(ac-bd\right)\left(ad-bc\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}ac-bd=0\\ad-bc=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}ac=bd\\ad=bc\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{a}{b}=\dfrac{d}{c}\\\dfrac{a}{b}=\dfrac{c}{d}\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}\dfrac{a}{b}=\dfrac{d}{c}\\\dfrac{a}{b}=\dfrac{c}{d}\end{matrix}\right.\) (ĐPCM)
12. Ta có \(ab\le\frac{a^2+b^2}{2}\)
=> \(a^2-ab+3b^2+1\ge\frac{a^2}{2}+\frac{5}{2}b^2+1\)
Lại có \(\left(\frac{a^2}{2}+\frac{5}{2}b^2+1\right)\left(\frac{1}{2}+\frac{5}{2}+1\right)\ge\left(\frac{a}{2}+\frac{5}{2}b+1\right)^2\)
=> \(\sqrt{a^2-ab+3b^2+1}\ge\frac{a}{4}+\frac{5b}{4}+\frac{1}{2}\)
=> \(\frac{1}{\sqrt{a^2-ab+3b^2+1}}\le\frac{4}{a+b+b+b+b+b+1+1}\le\frac{4}{64}.\left(\frac{1}{a}+\frac{5}{b}+2\right)\)
Khi đó
\(P\le\frac{1}{16}\left(6\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)+6\right)\le\frac{3}{2}\)
Dấu bằng xảy ra khi a=b=c=1
Vậy \(MaxP=\frac{3}{2}\)khi a=b=c=1
13. Ta có \(\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\le1\)
\(\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\ge\frac{9}{a+b+c+3}\)( BĐT cosi)
=> \(1\ge\frac{9}{a+b+c+3}\)
=> \(a+b+c\ge6\)
Ta có \(a^3-b^3=\left(a-b\right)\left(a^2+ab+b^2\right)\)
=> \(\frac{a^3-b^3}{a^2+ab+b^2}=a-b\)
Tương tự \(\frac{b^3-c^3}{b^2+bc+c^2}=b-c\),,\(\frac{c^3-a^2}{c^2+ac+a^2}=c-a\)
Cộng 3 BT trên ta có
\(\frac{a^3}{a^2+ab+b^2}+\frac{b^3}{b^2+bc+c^2}+\frac{c^3}{c^2+ac+c^2}=\frac{b^3}{a^2+ab+b^2}+\frac{c^3}{c^2+bc+b^2}+\frac{a^3}{a^2+ac+c^2}\)
Khi đó \(2P=\frac{a^3+b^3}{a^2+ab+b^2}+...\)
=> \(2P=\frac{\left(a+b\right)\left(a^2-ab+b^2\right)}{a^2+ab+b^2}+....\)
Xét \(\frac{a^2-ab+b^2}{a^2+ab+b^2}\ge\frac{1}{3}\)
<=> \(3\left(a^2-ab+b^2\right)\ge a^2+ab+b^2\)
<=> \(a^2+b^2\ge2ab\)(luôn đúng )
=> \(2P\ge\frac{1}{3}\left(a+b+b+c+a+c\right)=\frac{2}{3}.\left(a+b+c\right)\ge4\)
=> \(P\ge2\)
Vậy \(MinP=2\)khi a=b=c=2
Lưu ý : Chỗ .... là tương tự
Lời giải:
Đặt \(\left(\frac{ab}{c}, \frac{bc}{a}, \frac{ca}{b}\right)=(x,y,z)\)
Khi đó: \(xy=b^2; yz=c^2; xz=a^2\). Bài toán trở về dạng:
Cho $x,y,z>0$ thỏa mãn: \(xy+yz+xz=1\)
Tìm GTNN của \(P=x+y+z\)
Thật vậy: Ta đã biết một BĐT quen thuộc theo AM-GM là:
\((x+y+z)^2\geq 3(xy+yz+xz)\)
\(\Rightarrow x+y+z\geq \sqrt{3(xy+yz+xz)}=\sqrt{3}\)
Vậy \(P_{\min}=\sqrt{3}\)
Dấu bằng xảy ra khi \(x=y=z\Leftrightarrow a=b=c=\frac{1}{\sqrt{3}}\)