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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 \(a=b=c=1\)
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
Áp dụng bất đẳng thức \(AM-GM\) cho từng cặp số không âm, ta có:
\(a^2+b^2\ge2ab\) \(\left(1\right)\)
\(b^2+1\ge2b\) \(\left(2\right)\)
Cộng \(\left(1\right)\) và \(\left(2\right)\) vế theo vế, ta được:
\(a^2+2b^2+1\ge2ab+2b\)
\(\Rightarrow\) \(a^2+2b^2+3\ge2ab+2b+2\)
Vì hai vế của bất đẳng thức trên cùng dấu (do \(a,b,c>0\)) nên ta nghịch đảo hai vế và đổi chiều bất đẳng thức:
\(\frac{1}{a^2+2b^2+3}\le\frac{1}{2ab+2b+2}\) \(\left(1\right)\)
Hoàn toàn tương tự với vòng hoán vị \(b\) \(\rightarrow\) \(c\) \(\rightarrow\) \(a\) \(\rightarrow\) \(b\), ta có:
\(\frac{1}{b^2+2c^2+3}\ge\frac{1}{2bc+2c+2}\) \(\left(2\right)\) và \(\frac{1}{c^2+2a^2+3}\ge\frac{1}{2ca+2a+2}\) \(\left(3\right)\)
Cộng từng vế \(\left(1\right);\) \(\left(2\right)\) và \(\left(3\right)\), ta được:
\(VT\le\frac{1}{2ab+2b+2}+\frac{1}{2bc+2c+2}+\frac{1}{2ca+2a+2}=\frac{1}{2}\left(\frac{1}{ab+b+1}+\frac{1}{bc+c+1}+\frac{1}{ca+a+1}\right)\) \(\left(\text{*}\right)\)
Mặt khác, xét từng phân thức \(\frac{1}{ab+b+1};\frac{1}{bc+c+1};\frac{1}{ca+a+1}\) kết hợp với giả thiết đã cho, nghĩa là \(abc=1,\) ta có:
\(\frac{1}{ab+b+1};\) \(\frac{1}{bc+c+1}=\frac{abc}{bc+c+abc}=\frac{ab}{ab+b+1}\) và \(\frac{1}{ca+a+1}=\frac{abc}{ca+a+abc}=\frac{bc}{bc+c+1}=\frac{bc}{bc+c+abc}=\frac{b}{ab+b+1}\)
Do đó, \(\frac{1}{ab+b+1}+\frac{1}{bc+c+1}+\frac{1}{ca+a+1}=\frac{1}{ab+b+1}+\frac{ab}{ab+b+1}+\frac{b}{ab+b+1}=1\) \(\left(\text{**}\right)\)
Từ \(\left(\text{*}\right)\) và \(\left(\text{**}\right)\) suy ra \(\frac{1}{a^2+2b^2+3}+\frac{1}{b^2+2c^2+3}+\frac{1}{c^2+2a^2+3}\le\frac{1}{2}\)
Dấu \("="\) xảy ra \(\Leftrightarrow\) \(a=b=c\)
Áp dụng bđt Cauchy-Schwarz dạng Engel ta có:
a3/b+2c + b3/c+2a + c3/a+2b = a4/ab+2ac + b4/bc+2ab + c4/ac+2bc\(\ge\frac{\left(a^2+b^2+c^2\right)^2}{3\left(ab+bc+ca\right)}=\frac{1}{3\left(ab+bc+ca\right)}\)\(\ge\frac{1}{3\left(a^2+b^2+c^2\right)}=\frac{1}{3}\left(ĐPCM\right)\)
Do \(a,b< 1\Rightarrow a^3< a^2< a< 1;b^3< b^2< b< 1\)Ta có:\(\left(1-a^2\right)\left(1-b\right)>0\Rightarrow1+a^2b>a^2b\)
\(\Rightarrow1+a^2b>a^3+b^3haya^3+b^3< 1+a^2b\)Tương tự \(b^3+c^3< 1+b^2c;c^3+a^3< 1+c^2a\)
\(\Rightarrow2a^3+2b^3+2c^3< 3+a^2b+b^2c+c^2a\)
VT=2a2b2+2a2c2+2b2c2-a4-b4-c4
=a2b2+a2c2+b2c2+a2.(b2-a2)+b2.(c2-b2)+c2.(a2-c2)
=a2b2+a2c2+b2c2+a2.(b+a)(b-a)+b2.(c+b)(c-b)+c2.(a+c)(a-c)
Ta lại có : a+b>c=>a-c>-b
b+c>a=>b-a>-c
c+a>b=>c-b>-a
(BĐT tam giác)
=>VT>a2b2+a2c2+b2c2+a2.c.(-c)+b2.a.(-a)+c2.b.(-b)
=0
=>VT>0 =>dpcm
\(\frac{\left(2-c\right)\left(b-c\right)}{2a+bc}=\frac{\left(a+b\right)\left(b-c\right)}{a\left(a+b+c\right)+bc}=\frac{\left(a+b\right)\left(b-c\right)}{\left(a+b\right)\left(c+a\right)}=\frac{b-c}{c+a}=\frac{b}{c+a}-\frac{c}{c+a}\)
Tương tự, ta có: \(\frac{\left(2-a\right)\left(c-a\right)}{2b+ca}=\frac{c}{a+b}-\frac{a}{a+b};\frac{\left(2-b\right)\left(a-b\right)}{2c+ab}=\frac{a}{b+c}-\frac{b}{b+c}\)
\(\Rightarrow\)\(VT=\left(\frac{a}{b+c}-\frac{a}{a+b}\right)+\left(\frac{b}{c+a}-\frac{b}{b+c}\right)+\left(\frac{c}{a+b}-\frac{c}{c+a}\right)\)
\(=\frac{a\left(a-c\right)}{\left(a+b\right)\left(b+c\right)}+\frac{b\left(b-a\right)}{\left(b+c\right)\left(c+a\right)}+\frac{c\left(c-b\right)}{\left(c+a\right)\left(a+b\right)}\)
\(=\frac{a\left(a-c\right)\left(c+a\right)+b\left(b-a\right)\left(a+b\right)+c\left(c-b\right)\left(b+c\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)
\(=\frac{\left(a^3+b^3+c^3\right)-\left(a^2b+b^2c+c^2a\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\ge\frac{\left(a^3+b^3+c^3\right)-\left(a^3+b^3+c^3\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}=0\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(a=b=c=\frac{2}{3}\)
cái bđt \(a^3+b^3+c^3\ge a^2b+b^2c+c^2a\) cô Chi có làm r ib mk gửi link
Ta có: \(1=a^2+b^2+c^2\ge ab+bc+ca\).
\(P=\dfrac{a^3}{b+2c}+\dfrac{b^3}{c+2a}+\dfrac{c^3}{a+2b}=\dfrac{a^4}{ab+2ca}+\dfrac{b^4}{bc+2ab}+\dfrac{c^4}{ca+2bc}\)
\(\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{3\left(ab+bc+ca\right)}=\dfrac{1}{3\left(ab+bc+ca\right)}\ge\dfrac{1}{3}\)
Dấu \(=\) xảy ra khi \(a=b=c=\dfrac{1}{\sqrt{3}}\).
Ta có:\(a^2+2b+3=a^2+2b+1+2\ge2\left(a+b+1\right)\)
Tương tự ta được:\(VT\le\frac{1}{2}\left(\frac{a}{a+b+1}+\frac{b}{b+c+1}+\frac{c}{c+a+1}\right)\)
Ta sẽ chứng minh \(\frac{a}{a+b+1}+\frac{b}{b+c+1}+\frac{c}{c+a+1}\le1\)
\(\Leftrightarrow\frac{-b-1}{a+b+1}+\frac{-c-1}{b+c+1}+\frac{-a-1}{c+a+1}\le-2\)
\(\Leftrightarrow\frac{b+1}{a+b+1}+\frac{c+1}{b+c+1}+\frac{a+1}{c+a+1}\ge2\)
\(\Leftrightarrow\frac{\left(b+1\right)^2}{\left(b+1\right)\left(a+b+1\right)}+\frac{\left(c+1\right)^2}{\left(c+1\right)\left(b+c+1\right)}+\frac{\left(a+1\right)^2}{\left(a+1\right)\left(c+a+1\right)}\ge2\)(*)
Áp dụng Bđt Cauchy-Schwarz dạng engel ta có:
VT(*)\(\ge\frac{\left(a+b+c+3\right)^2}{a^2+b^2+c^2+ab+bc+ca+3\left(a+b+c\right)+3}\)
Mà \(a^2+b^2+c^2+ab+bc+ca+3\left(a+b+c\right)+3\)
\(=\frac{1}{2}\left[a^2+b^2+c^2+2\left(ab+bc+ca\right)+6\left(a+b+c\right)+9\right]\)
\(=\frac{1}{2}\left(a+b+c+3\right)^2\)
=>VT(*)\(\ge\)2=VP (*)
Vậy Bđt được chứng minh
Cho hỏi VT;VP là gì