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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\(\ge\frac{\left(a^2+b^2+c^2\right)^2}{3\left(ab+bc+ca\right)}=\frac{1}{3\left(ab+bc+ca\right)}\)\(\ge\frac{1}{3\left(a^2+b^2+c^2\right)}=\frac{1}{3}\left(ĐPCM\right)\)
Lời giải:
Áp dụng BĐT Bunhiacopkxy:
\((2a^2+b^2)(2a^2+c^2)=(a^2+a^2+b^2)(a^2+c^2+a^2)\geq (a^2+ac+ab)^2\)
\(=[a(a+b+c)]^2\)
\(\Rightarrow \frac{a^3}{(2a^2+b^2)(2a^2+c^2)}\leq \frac{a^3}{[a(a+b+c)]^2}=\frac{a}{(a+b+c)^2}\)
Hoàn toàn tương tự với các phân thức còn lại và cộng theo vế thu được:
\(\sum \frac{a^3}{(2a^2+b^2)(2a^2+c^2)}\leq \frac{a+b+c}{(a+b+c)^2}=\frac{1}{a+b+c}\) (đpcm)
Dấu "=" xảy ra khi $a=b=c$
Áp dụng BĐT Cô-si ta có:
\(a^2+b^2\ge2ab;b^2+1^2\ge2b\)
\(\Rightarrow a^2+b^2+b^2+1+2\ge2ab+2b+2\)
\(\Rightarrow a^2+2b^2+3\ge2\left(ab+b+1\right)\)
\(\Rightarrow\frac{1}{a^2+2b^2+3}\le\frac{1}{2\left(ab+b+1\right)}=\frac{1}{2}.\frac{1}{ab+b+1}\)
chứng minh tương tự
\(\Rightarrow\frac{1}{b^2+2c^2+3}\le\frac{1}{2}.\frac{1}{bc+c+1};\frac{1}{c^2+2a^2+3}\le\frac{1}{2}.\frac{1}{ac+a+1}\)
\(\Rightarrow\frac{1}{a^2+2b^2+3}+\frac{1}{b^2+2c^2+3}+\frac{1}{c^2+2a^2+3}\le\frac{1}{2}.\frac{1}{ab+b+1}+\frac{1}{2}.\frac{1}{bc+c+1}+\frac{1}{2}.\frac{1}{ac+a+1}\)
\(\Rightarrow\frac{1}{a^2+2b^2+3}+\frac{1}{b^2+2c^2+3}+\frac{1}{c^2+2a^2+3}\le\frac{1}{2}.\left(\frac{1}{ab+b+1}+\frac{1}{bc+c+1}+\frac{1}{ac+a+1}\right)\)
đặt \(A=\frac{1}{ab+b+1}+\frac{1}{bc+c+1}+\frac{1}{ac+a+1}\)
\(=\frac{ac}{a^2bc+abc+ac}+\frac{a}{abc+ac+a}+\frac{1}{ac+a+1}\)
\(=\frac{ac}{ac+a+1}+\frac{a}{ac+a+1}+\frac{1}{ac+a+1}=\frac{ac+a+1}{ac+a+1}=1\)
\(\Rightarrow\frac{1}{a^2+2b^2+3}+\frac{1}{b^2+2c^2+3}+\frac{1}{c^2+2a^2+3}\le\frac{1}{2}.1=2\)
=>đpcm
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{a}+\frac{1}{c}+\frac{1}{b}+\frac{1}{c}\ge4\left(\frac{1}{a+b}+\frac{1}{a+c}+\frac{1}{b+c}\right)\ge2\)
\(\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge1\)
Đặt \(\left(\frac{1}{a};\frac{1}{b};\frac{1}{c}\right)=\left(x;y;z\right)\Rightarrow x+y+z\ge1\)
\(P=\sqrt{x^2+2y^2}+\sqrt{y^2+2z^2}+\sqrt{z^2+2x^2}\)
\(\Rightarrow P\ge\sqrt{\frac{\left(x+2y\right)^2}{3}}+\sqrt{\frac{\left(y+2z\right)^2}{3}}+\sqrt{\frac{\left(z+2x\right)^2}{3}}\)
\(\Rightarrow P\ge\frac{1}{\sqrt{3}}\left(3x+3y+3z\right)\ge\frac{3}{\sqrt{3}}=\sqrt{3}\)
Dấu "=" xảy ra khi \(x=y=z=\frac{1}{3}\) hay \(a=b=c=3\)
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
1) Áp dụng bunhiacopxki ta được \(\sqrt{\left(2a^2+b^2\right)\left(2a^2+c^2\right)}\ge\sqrt{\left(2a^2+bc\right)^2}=2a^2+bc\), tương tự với các mẫu ta được vế trái \(\le\frac{a^2}{2a^2+bc}+\frac{b^2}{2b^2+ac}+\frac{c^2}{2c^2+ab}\le1< =>\)\(1-\frac{bc}{2a^2+bc}+1-\frac{ac}{2b^2+ac}+1-\frac{ab}{2c^2+ab}\le2< =>\)
\(\frac{bc}{2a^2+bc}+\frac{ac}{2b^2+ac}+\frac{ab}{2c^2+ab}\ge1\)<=> \(\frac{b^2c^2}{2a^2bc+b^2c^2}+\frac{a^2c^2}{2b^2ac+a^2c^2}+\frac{a^2b^2}{2c^2ab+a^2b^2}\ge1\) (1)
áp dụng (x2 +y2 +z2)(m2+n2+p2) \(\ge\left(xm+yn+zp\right)^2\)
(2a2bc +b2c2 + 2b2ac+a2c2 + 2c2ab+a2b2). VT\(\ge\left(bc+ca+ab\right)^2\) <=> (ab+bc+ca)2. VT \(\ge\left(ab+bc+ca\right)^2< =>VT\ge1\) ( vậy (1) đúng)
dấu '=' khi a=b=c
Ta có: \(a^2+b^2\ge2ab;b^2+1\ge2b\Rightarrow a^2+2b^2+3\ge2\left(ab+b+1\right)\)
\(\Rightarrow\frac{1}{a^3+2b^2+3}\le\frac{1}{2\left(ab+b+1\right)}\)
Tương tự ta cũng có:
\(\frac{1}{a^2+2b^2+3}+\frac{1}{b^2+2c^2+3}+\frac{1}{c^2+2a^2+3}\le\)\(\frac{1}{2}\left(\frac{1}{ab+b+1}+\frac{1}{bc+b+1}+\frac{1}{ca+a+1}\right)\)
Mà: \(\frac{1}{ab+b+1}+\frac{1}{bc+c+1}+\frac{1}{ca+a+1}=\frac{1}{ab+b+1}+\)\(\frac{ab}{ab^2+abc+ab}+\frac{b}{bca+ab+b}=1\)
\(\Rightarrow\frac{1}{a^2+2b^2+3}+\frac{1}{b^2+2c^2+3}+\frac{1}{c^2+2a^2+3}\le\frac{1}{2}\left(đpcm\right)\)\(\Leftrightarrow a=b=c=1\)
Đề sao rồi bạn ơi, phải là \(\le\) mới đúng. Bài này ta làm như sau:
Áp dụng BĐT Cauchy, ta có:
\(a^2+2b^2+3=\left(a^2+b^2\right)+\left(b^2+1\right)+2\ge2ab+2b+2=2\left(ab+b+1\right)\)
CMTT, ta được:
\(b^2+2c^2+3\ge2\left(bc+c+1\right)\)
\(c^2+2a^2+3\ge2\left(ca+a+1\right)\)
Do đó ta có:
\(\frac{1}{a^2+2b^2+3}+\frac{1}{b^2+2c^2+3}+\frac{1}{c^2+2a^2+3}\le\frac{1}{2}\left(\frac{1}{ab+b+1}+\frac{1}{bc+c+1}+\frac{1}{ca+a+1}\right)\left(1\right)\)
Chú ý rằng \(abc=1\) nên ta dễ dàng CM được:
\(\frac{1}{ab+b+1}+\frac{1}{bc+c+1}+\frac{1}{ca+a+1}=1\left(2\right)\)
Từ \(\left(1\right)\) và \(\left(2\right)\) ta có đpcm.
Nếu không cho abc=1; a,b,c >0 và BĐT >=1 thì mình xong lâu rồi. Khó phết