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Áp dụng bất đẳng thức Cô si cho hai số dương ta có:
(a2 + b2) + (b2 + c2) + (c2 + a2) ≥ 2ab + 2bc + 2ca
=> 2(a2 + b2 + c2 ) ≥ 2 (ab + bc + ca) (1) (a2 + 1) + (b2 + c2) + (c2 + a2) ≥ 2a + 2b + 2c
=> a2 + b2 + c2 + 3 ≥ 2(a + b + c) (2)
Cộng các vế của (1) và (2) ta có:
3 ( a2 + b2 + c2 ) + 3 ≥ 2 (ab + bc + ca + a + b + c)
=> 3( a2 + b2 + c2 ) + 3 ≥ 12 => a2 + b2 + c2 ≥ 3.
Ta có: (a^3/b + ab ) + ( b^3/c + bc ) + ( c^3/a + ca)≥ 2(a2 + b2 + c2) (CÔ SI)
<=>a^3/b + b^3/c + c^3/a +ab + bc + ac ≥ 2(a2 + b2 + c2)
Vì a2 + b2 + c2 ≥ ab + bc + ca => a^3 + b^3 + c^3 ≥ a2 + b2 + c2 ≥ 3 (đpcm).
Áp dụng bất đẳng thức cô-si cho hai số dương ta có:
\(\left(a^2+b^2\right)+\left(b^2+c^2\right)+\left(c^2+a^2\right)\ge2ab+2bc+2ca\)
\(\Rightarrow2\left(a^2+b^2+c^2\right)\ge2\left(ab+bc+ca\right)\) (1)
\(\left(a^2+b^2\right)+\left(b^2+c^2\right)+\left(c^2+a^2\right)\ge2a+2b+2c\)
\(\Rightarrow a^2+b^2+c^2+3\ge2\left(a+b+c\right)\) (2)
Cộng (1) với (2)
\(3\left(a^2+b^2+c^2\right)+3\ge2\left(ab+bc+ca+a+b+c\right)\)
\(\Rightarrow3\left(a^2+b^2+c^2\right)+3\ge12\)
\(\Rightarrow a^2+b^2+c^2\ge3\)
Ta có: \(\left(\dfrac{a^3}{b}+ab\right)+\left(\dfrac{b^3}{c}+bc\right)+\left(\dfrac{c^3}{a}+ca\right)\ge2\left(a^2+b^2+c^2\right)\)
\(\Leftrightarrow\dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a}+ab+bc+ca\ge2\left(a^2+b^2+c^2\right)\)
Vì \(a^2+b^2+c^2\ge ab+bc+ca\)
\(\Rightarrow\dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a}\ge a^2+b^2+c^2\ge3\) (đpcm).
Áp dụng bất đẳng thức Cauchy
\(\Rightarrow VT\ge3\sqrt[6]{\dfrac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{\left(c+ab\right)\left(a+bc\right)\left(b+ac\right)}}\)
Chứng minh \(3\sqrt[6]{\dfrac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{\left(c+ab\right)\left(a+bc\right)\left(b+ac\right)}}\ge3\)
\(\Leftrightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge\left(c+ab\right)\left(a+bc\right)\left(b+ac\right)\)
Áp dụng bất đẳng thức Cauchy
\(\Rightarrow\left(c+ab\right)\left(a+bc\right)\le\dfrac{\left(c+a+ab+bc\right)^2}{4}=\dfrac{\left[b\left(a+c\right)+c+a\right]^2}{4}=\dfrac{\left(b+1\right)^2\left(c+a\right)^2}{4}\)
Thiết lập tương tự và thu lại ta có
\(\Rightarrow\left(c+ab\right)^2\left(a+bc\right)^2\left(b+ac\right)^2\le\dfrac{\left(a+b\right)^2\left(b+c\right)^2\left(c+a\right)^2\left(b+1\right)^2\left(a+1\right)^2\left(c+1\right)^2}{64}\)
\(\Rightarrow64\left(c+ab\right)^2\left(a+bc\right)^2\left(b+ac\right)^2\le\left(a+b\right)^2\left(b+c\right)^2\left(c+a\right)^2\left(b+1\right)^2\left(c+1\right)^2\left(a+1\right)^2\)
\(\Leftrightarrow8\left(c+ab\right)\left(a+bc\right)\left(b+ac\right)\le\left(a+b\right)\left(b+c\right)\left(c+a\right)\left(b+1\right)\left(c+1\right)\left(a+1\right)\)
Cần chứng minh rằng \(\left(a+1\right)\left(b+1\right)\left(c+1\right)\le8\)
Áp dụng bất đẳng thức Cauchy \(\left(a+1\right)\left(b+1\right)\left(c+1\right)\le\left(\dfrac{3+3}{3}\right)^3=8\left(đpcm\right)\)
\(\Rightarrowđpcm\)
Đặt\(P=\dfrac{a}{1+b^2}+\dfrac{b}{1+c^2}+\dfrac{c}{1+a^2+}+\dfrac{1}{2}\left(ab+bc+ca\right)\)
Bổ đề: \(\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge9\) \(\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{9}{a+b+c}\) (1)
Chứng minh bổ đề: \(\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge9\sqrt[3]{abc.\dfrac{1}{abc}}=9\left(\forall a,b,c\ge0\right)\)
Kết hợp điều kiện đề bài ta được: \(a+b+c\ge3\)
Ta có: \(\dfrac{ab^2}{1+b^2}\le\dfrac{ab^2}{2\sqrt{b^2}}=\dfrac{ab}{2}\) ( AM-GM cho 2 số không âm 1 và b^2 )
\(\Rightarrow\dfrac{a}{1+b^2}=a-\dfrac{ab^2}{1+b^2}\ge a-\dfrac{ab}{2}\left(1\right)\)
Chứng minh hoàn toàn tương tự: \(\dfrac{b}{1+c^2}\ge b-\dfrac{bc}{2}\left(2\right)\)
\(\dfrac{c}{1+a^2}\ge c-\dfrac{ca}{2}\left(3\right)\)
Cộng (1),(2),(3) vế theo vế thu được: \(P\ge a+b+c=3\)
Dấu "=" xảy ra tại a=b=c=1
iải của
\(BDT\Leftrightarrow2a^4b+2b^4c+2c^4a+3ab^4+3bc^4+3ca^4\ge5a^2b^2c+5a^2bc^2+5ab^2c^2\)
Ta chứng minh được \(ab^4+bc^4+ca^4\ge a^2b^2c+a^2bc^2+ab^2c^2\)
\(\Leftrightarrow\dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a}\ge ab+bc+ca\)
\(VT=\dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a}=\dfrac{a^4}{ab}+\dfrac{b^4}{bc}+\dfrac{c^4}{ac}\)
\(\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{ab+bc+ca}\ge\dfrac{\left(ab+bc+ca\right)^2}{ab+bc+ca}=VP\)
Vậy ta cần chứng minh \(2a^4b+2b^4c+2c^4a+2ab^4+2bc^4+2ca^4\ge4a^2b^2c+4a^2bc^2+4ab^2c^2\)
\(\Leftrightarrow\sum_{cyc}\left(2c^3+bc^2-b^2c+ac^2-a^2c+3ab^2+3a^2b\right)\left(a-b\right)^2\ge0\)
Dấu "=" xảy ra khi \(a=b=c\)
Ta có:
\(\left(a^2+1\right)+\left(b^2+1\right)+\left(c^2+1\right)+\left(a^2+b^2\right)+\left(b^2+c^2\right)+\left(c^2+a^2\right)\)
\(\ge2a+2b+2c+2ab+2bc+2ca=12\)
\(\Rightarrow3\left(a^2+b^2+c^2\right)+3\ge12\)
\(\Rightarrow a^2+b^2+c^2\ge3\)
\(P=\dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a}=\dfrac{a^4}{ab}+\dfrac{b^4}{bc}+\dfrac{c^4}{ca}\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{ab+bc+ca}\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{a^2+b^2+c^2}\)
\(P\ge a^2+b^2+c^2\ge3\)
\(P_{min}=3\) khi \(a=b=c=1\)
Ta có: \(\dfrac{a^3}{bc}+\dfrac{b^3}{ca}+\dfrac{c^3}{ab}+2a+2b+2c\)
\(=\left(\dfrac{a^3}{bc}+b+c\right)+\left(\dfrac{b^3}{ca}+a+c\right)+\left(\dfrac{c^3}{ab}+a+b\right)\ge3\sqrt[3]{\dfrac{a^3}{bc}.b.c}+3\sqrt[3]{\dfrac{b^3}{ca}.a.c}+3\sqrt[3]{\dfrac{c^3}{ab}.a.b}=3a+3b+3c\)
\(\Rightarrow\dfrac{a^3}{bc}+\dfrac{b^3}{ca}+\dfrac{c^3}{ab}+2a+2b+2c\ge3a+3b+3c\)
\(\Rightarrow\dfrac{a^3}{bc}+\dfrac{b^3}{ca}+\dfrac{c^3}{ab}\ge a+b+c\)
\("="\Leftrightarrow a=b=c\)
Ta có: \(A=\dfrac{a^3}{bc}+\dfrac{b^3}{ca}+\dfrac{c^3}{ab}=\dfrac{a^4}{abc}+\dfrac{b^4}{abc}+\dfrac{c^4}{abc}\)(do \(a;b;c>0\) )
Áp dụng BĐT \(a^2+b^2+c^2\ge ab+bc+ca\)(\("="\Leftrightarrow a=b=c\))
\(A=\dfrac{a^4+b^4+c^4}{abc}=\dfrac{\left(a^2\right)^2+\left(b^2\right)^2+\left(c^2\right)^2}{abc}\ge\)
\(\ge\dfrac{a^2b^2+b^2c^2+c^2a^2}{abc}\ge\dfrac{abc\left(a+b+c\right)}{abc}=a+b+c\)
https://hoc24.vn/hoi-dap/tim-kiem?q=Cho+c%C3%A1c+s%E1%BB%91+th%E1%BB%B1c+d%C6%B0%C6%A1ng+a,+b,+c+tho%E1%BA%A3+m%C3%A3n:+abc+a+b=3ababc+a+b=3ababc+a+b=3ab.+Ch%E1%BB%A9ng+minh+r%E1%BA%B1ng:+%E2%88%9Aaba+b+1+%E2%88%9Abbc+c+1+%E2%88%9Aaca+c+1%E2%89%A5%E2%88%9A3aba+b+1+bbc+c+1+aca+c+1%E2%89%A53\sqrt{\dfrac{ab}{a+b+1}}+\sqrt{\dfrac{b}{bc+c+1}}+\sqrt{\dfrac{a}{ca+c+1}}\ge\sqrt{3}&id=695796
Cách 1
Áp dụng bđt Cauchy ta có
\(\frac{a^3}{b}+b+1\ge3a,\frac{b^3}{c}+c+1\ge3b,\frac{c^3}{a}+a+1\ge3a\)
Cộng từng vế 3 bđt trên ta có
\(A=\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}\ge2\left(a+b+c\right)-3\)
Mặt khác (a+b+c)2+3(a+b+c)\(\ge\)18 (biến đổi tương đương là c/m được)
Đặt m=a+b+c
=> t2+3t-18\(\ge\)0
=> t\(\ge\)3
=> A\(\ge\)3
Dấu "=" xảy ra khi a=b=c=1
Cách 2,rất phức tạp :(
\(6=a+b+c+ab+bc+ca\le\frac{\left(a+b+c\right)^2+3\left(a+b+c\right)}{3}\)
Suy ra \(\left(a+b+c\right)^2+3\left(a+b+c\right)-18\ge0\)
\(\Leftrightarrow a+b+c\ge3\Leftrightarrow a^2+b^2+c^2+2\left(ab+bc+ca\right)\ge9\).
Mà \(VT\le3\left(a^2+b^2+c^2\right)\Rightarrow3\left(a^2+b^2+c^2\right)\ge9\Leftrightarrow a^2+b^2+c^2\ge3\)
Ta chứng minh BĐT sau = sos cho đẹp: \(\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}\ge a^2+b^2+c^2\)
\(\Leftrightarrow\Sigma_{cyc}\left(\frac{a^3}{b}-\frac{a^2b}{b}\right)\ge0\Leftrightarrow\Sigma_{cyc}\frac{a^2\left(a-b\right)}{b}-\Sigma_{cyc}a\left(a-b\right)+\Sigma_{cyc}a\left(a-b\right)\ge0\)
\(\Leftrightarrow\Sigma_{cyc}\frac{a^2\left(a-b\right)^2}{b}+\left(a^2+b^2+c^2-ab-bc-ca\right)\ge0\)
\(\Leftrightarrow\Sigma_{cyc}\frac{a^2\left(a-b\right)^2}{b}+\frac{1}{2}\left(a-b\right)^2\ge0\Leftrightarrow\left(a-b\right)^2\left(\frac{a^2}{b}+\frac{1}{2}\right)\ge0\) (đúng)
Do vậy: \(\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}\ge a^2+b^2+c^2\ge3^{\left(đpcm\right)}\)
Xảy ra đẳng thức khi a = b = c = 1
trời ơi , bài sai nhiều chỗ qá ....
Ơ , CÁI GÌ ZẬY TRỜI ! BÀI NÀY ĐỀ SAI MÀ TAO GIẢI ĐC ???