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Cho a + b + c = 3. Chứng minh \(\frac{a}{b^2+1}+\frac{b}{c^2+1}+\frac{c}{a^2+1}\)lớn hơn hoặc bằng 3
ta có \(a^2+2b^2+3=a^2+b^2+b^2+1+2.\)
áp dụng BĐT cauchy
=>\(a^2+2b^2+3>=2ab+2b+2=2\left(ab+b+1\right)\)
=>\(\frac{1}{a^2+2b^2+3}< =\frac{1}{2\left(ab+b+1\right)}\)
tương tự ta có \(\hept{\frac{1}{b^2+2c^2+3}< =\frac{1}{2\left(bc+c+1\right)}}\),\(\frac{1}{c^2+2a^2+3}< =\frac{1}{2\left(ac+a+1\right)}\)
=>VT<=\(\frac{1}{2}.\left(\frac{1}{ab+b+1}+\frac{1}{ac+a+1}+\frac{1}{bc+c+1}\right)\)
<=>VT<=\(\frac{1}{2}\left(\frac{1}{ab+b+1}+\frac{abc}{ac+a^2bc+abc}+\frac{abc}{bc+c+abc}\right)\)(do abc=1)
<=>VT<=\(\frac{1}{2}\left(\frac{1}{ab+b+1}+\frac{b}{ab+b+1}+\frac{ab}{ab+b+1}\right)\)=\(\frac{1}{2}\left(\frac{ab+b+1}{ab+b+1}\right)=\frac{1}{2}\)(đpcm)
Dấu bằng xảy ra khi a=b=c=1
1/(a^2+2b^2+3)+1/(b^2+2c^2+3)+1/(c^2+2a^2+3)
Tại có: abc=1 =>a=1;b=1;c=1.
Syu ra: 1/(1+2.1+3)+1/(1+2.1+3)+1/(1+2.1+3)
=1/6+1/6+1/6=1/2
=>1/(a^2+2b^2+3)+1/(b^2+2c^2+3)+1/(c^2+2a^2+3) \(\le\)1/2
=> đpcm
\(a^3+b^3+c^3-3abc=0\)
\(\Leftrightarrow\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc=0\)
\(\Leftrightarrow\left(a+b+c\right)\left[\left(a+b\right)^2-c\left(a+b\right)+c^2\right]-3ab\left(a+b+c\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(2a^2+2b^2+2c^2-2ab-2bc-2ca\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a+b+c=0\\\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}a+b+c=0\\a=b=c\end{matrix}\right.\)
Áp dụng: \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\Rightarrow\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{3}{abc}\)
\(A=abc\left(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}\right)=abc.\frac{3}{abc}=3\)
a) a2+b2-2ab=(a-b)2>=0
b) \(\frac{a^2+b^2}{2}\)\(\ge\)ab <=> \(\frac{a^2+b^2}{2}\)-ab\(\ge\)0 <=> \(\frac{\left(a-b\right)^2}{2}\)\(\ge\)0 (ĐPCM)
c) a2+2a < (a+1)2=a2+2a+1 (ĐPCM)
Ta có \(1=a+b+c\ge3\sqrt[3]{abc}\)
\(\Leftrightarrow\frac{1}{3}\ge\sqrt[3]{abc}\)
Theo đề bài ta có
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{ab+bc+ca}{abc}\)
\(\ge\frac{3\sqrt[3]{a^2b^2c^2}}{abc}=\frac{3}{\sqrt[3]{abc}}\ge9\)
Nhân cả 2 vế với a+b+c
Chứng minh \(\frac{a}{b}+\frac{b}{a}\ge2\) tương tự với \(\frac{b}{c}+\frac{c}{b};\frac{c}{a}+\frac{a}{c}\)
\(\Leftrightarrow\frac{a}{b}+\frac{b}{a}-2\ge0\Leftrightarrow\frac{a^2-2ab+b^2}{ab}\ge0\Leftrightarrow\frac{\left(a-b\right)^2}{ab}\ge0\)luôn đúng do a;b>0
dễ rồi nhé
b) \(P=\frac{x}{x+1}+\frac{y}{y+1}+\frac{z}{z+1}\)
\(P=\left(\frac{x+1}{x+1}+\frac{y+1}{y+1}+\frac{z+1}{z+1}\right)-\left(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\right)\)
\(P=\left(1+1+1\right)-\left(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\right)\)
\(P=3-\left(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\right)\)
Áp dụng bđt Cauchy Schwarz dạng Engel (mình nói bđt như vậy,chỗ này bạn cứ nói theo cái bđt đề bài cho đi) ta được:
\(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\ge\frac{\left(1+1+1\right)^2}{x+1+y+1+z+1}=\frac{9}{4}\)
=>\(P=3-\left(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\right)\le3-\frac{9}{4}=\frac{3}{4}\)
=>Pmax=3/4 <=> x=y=z=1/3