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\(0< a;b;c< 1\Rightarrow\left\{{}\begin{matrix}a^3< a^2< a< 1\\b^3< b^2< b< 1\\c^3< c^2< c< 1\end{matrix}\right.\)
\(\Rightarrow\left(1-a\right)\left(1-b^2\right)>0\Rightarrow1+ab^2>a+b^2>a^3+b^3\)
Tương tự: \(1+b^2c>b^3+c^3\); \(1+ca^2>a^3+c^3\)
Cộng vế với vế: \(3+a^2b+b^2c+c^2a>2a^3+2b^3+2c^3\)
Đẳng thức không xảy ra
Áp dụng BĐT AM - GM, ta có:
\(a^2+2b^2+3\)
\(=\left(a^2+b^2\right)+\left(b^2+1\right)+2\)
\(\ge2ab+2b+2\)
Tương tự, ta có: \(b^2+2c^2+3\ge2bc+2c+2\) và \(c^2+2a^2+3\ge2ac+2a+2\)
\(VT=\dfrac{1}{a^2+2b^2+3}+\dfrac{1}{b^2+2c^2+3}+\dfrac{1}{c^2+2a^2+3}\)
\(\le\dfrac{1}{2ab+2b+2}+\dfrac{1}{2bc+2c+2}+\dfrac{1}{2ac+2a+2}\)
\(=\dfrac{1}{2}\left(\dfrac{1}{ab+b+1}+\dfrac{1}{bc+c+1}+\dfrac{1}{ac+a+1}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{ab+b+1}+\dfrac{abc}{bc+c+abc}+\dfrac{abc}{ac+a^2bc+abc}\right)\) (Thay abc = 1)
\(=\dfrac{1}{2}\left(\dfrac{1}{ab+b+1}+\dfrac{ab}{b+1+ab}+\dfrac{b}{1+ab+b}\right)\)
\(=\dfrac{1}{2}\times\dfrac{1+ab+b}{ab+b+1}\)
\(=\dfrac{1}{2}=VP\left(\text{đ}pcm\right)\)
Dấu "=" xảy ra khi a = b = c = 1
Bao nhiêu công gõ bài xong rồi đi chơi, chơi về định gửi bài, chơi về bật máy lên gửi thì lỗi, may vãi
Ta có:
\(\dfrac{a^2}{\left(2a+b\right)\left(2a+c\right)}=\dfrac{a^2}{2a\left(a+b+c\right)+2a^2+bc}\)
\(\le\dfrac{1}{9}\left(\dfrac{a^2}{a\left(a+b+c\right)}+\dfrac{a^2}{a\left(a+b+c\right)}+\dfrac{a^2}{2a^2+bc}\right)\)
\(=\dfrac{1}{9}\left(\dfrac{2a}{a+b+c}+\dfrac{a^2}{2a^2+bc}\right)\)
Tương tự cho 2 BĐT còn lại rồi cộng theo vế:
\(VT\le\dfrac{1}{9}\left(\dfrac{2\left(a+b+c\right)}{a+b+c}+\dfrac{a^2}{2a^2+bc}+\dfrac{b^2}{2b^2+ac}+\dfrac{c^2}{2c^2+ab}\right)\)
\(=\dfrac{1}{9}\left(2+\dfrac{a^2}{2a^2+bc}+\dfrac{b^2}{2b^2+ac}+\dfrac{c^2}{2c^2+ab}\right)\)
Cần chứng minh \(\dfrac{1}{9}\left(2+\dfrac{a^2}{2a^2+bc}+\dfrac{b^2}{2b^2+ac}+\dfrac{c^2}{2c^2+ab}\right)\le\dfrac{1}{3}\)
\(\Leftrightarrow\dfrac{a^2}{2a^2+bc}+\dfrac{b^2}{2b^2+ac}+\dfrac{c^2}{2c^2+ab}\le1\)
\(\Leftrightarrow\dfrac{bc}{bc+2a^2}+\dfrac{ca}{ca+2b^2}+\dfrac{ab}{ab+2c^2}\ge1\)
Cauchy-Schwarz: \(VT=\dfrac{bc}{bc+2a^2}+\dfrac{ca}{ca+2b^2}+\dfrac{ab}{ab+2c^2}\)
\(=\dfrac{b^2c^2}{b^2c^2+2a^2bc}+\dfrac{c^2a^2}{c^2a^2+2ab^2c}+\dfrac{a^2b^2}{a^2b^2+2abc^2}\)
\(\ge\dfrac{\left(ab+bc+ca\right)^2}{\left(ab+bc+ca\right)^2}=1\) * Đúng*
Happy New Year (Lunar)
\(abc\le1\)
\(VT=\sum\dfrac{a^4}{2abc+a^2b}\ge\dfrac{\sum^2a^2}{6+\sum a^2b}\ge\dfrac{\sum^2a^2}{6+\sqrt{\dfrac{1}{3}\sum^3a^2}}\)
Ta cần chứng minh :
\(\dfrac{\sum^2a^2}{6+\sqrt{\dfrac{1}{3}\sum^3a^2}}\ge1\)
Đặt \(\sum a^2=t\left(t\ge3\right)\)
\(\Rightarrow\dfrac{t^2}{6+\sqrt{\dfrac{1}{3}t^3}}\ge1\Leftrightarrow t\sqrt{t}\left(\sqrt{t}-\dfrac{1}{\sqrt{3}}\right)\ge6\)
Thật vậy :
\(t\sqrt{t}\left(\sqrt{t}-\dfrac{1}{\sqrt{3}}\right)\ge3\sqrt{3}\left(\sqrt{3}-\dfrac{1}{\sqrt{3}}\right)=6\left(t\ge3\right)\)
Hôm qua em không có online. Bài này căng não@@
Đặt \(p=a+b+c;q=ab+bc+ca;r=abc\Rightarrow q=3\) thì \(p^2\ge3q=9\Rightarrow p\ge3\)
Chú ý: \(-4p^3r + p^2q^2 + 18pqr - 4q^3 - 27r^2=(a-b)^2 (b-c)^2 (c-a)^2 \geq 0\)
\(\Rightarrow\) \(1/27(-2p^3-2\sqrt{(p^2-3q)^3}+9pq) \leq r \leq 1/27(-2p^3+2\sqrt{(p^2-3q)^3}+9pq)\)
Hay là: \(\frac{1}{27}\left(-2p^3-2\sqrt{\left(p^2-9\right)^3}+27p\right)\le r\le\frac{1}{27}\left(-2p^3+2\sqrt{\left(p^2-9\right)^3}+27p\right)\)
Nếu \(a\ge b\ge c\Rightarrow a^2b+b^2c+c^2a\ge ab^2+bc^2+ca^2\)
\(\Rightarrow a^2b+b^2c+c^2a\ge\frac{1}{2}\Sigma ab\left(a+b\right)=\frac{1}{2}\left(pq-3r\right)=\frac{3}{2}\left(p-3r\right)\)
Do đó: \(P\ge\frac{1}{2}\left(p-3r\right)+\sqrt[3]{9p}\ge\frac{1}{2}\left(p-\frac{1}{27}\left(-2p^3+2\sqrt{\left(p^2-9\right)^3}+27p\right)\right)+3\)
\(\ge\frac{1}{27}p^3-\frac{1}{27}\sqrt{\left(p^2-9\right)^3}+3=f\left(p\right)\). Dễ thấy khi p tăng thì f(p) tăng.
Do đó f(p) đạt giá trị nhỏ nhất khi p đạt giá trị nhỏ nhất. Hay là: \(f\left(p\right)\ge f\left(3\right)=4=VP\)
Trường hợp còn lại tối về em đăng, đang bận!
Nếu \(a\le b\le c\Rightarrow\left(a-b\right)\left(b-c\right)\left(a-c\right)\le0\)
\(\Rightarrow\left(a-b\right)\left(b-c\right)\left(a-c\right)=-\left|\left(a-b\right)\left(b-c\right)\left(a-c\right)\right|=-\sqrt{\left(a-b\right)^2\left(b-c\right)^2\left(c-a\right)^2}\)
\(=-\sqrt{-4p^3r + p^2q^2 + 18pqr - 4q^3 - 27r^2}\)
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Chú ý: \(-4p^3r + p^2q^2 + 18pqr - 4q^3 - 27r^2=(a-b)^2 (b-c)^2 (c-a)^2 \geq 0\)
\(\Rightarrow\) \(1/27(-2p^3-2\sqrt{(p^2-3q)^3}+9pq) \leq r \leq 1/27(-2p^3+2\sqrt{(p^2-3q)^3}+9pq)\)
Hay là: \(\frac{1}{27}\left(-2p^3-2\sqrt{\left(p^2-9\right)^3}+27p\right)\le r\le\frac{1}{27}\left(-2p^3+2\sqrt{\left(p^2-9\right)^3}+27p\right)\)
Ta có: \(2\left(a^2b+b^2c+c^2a\right)=\Sigma ab\left(a+b\right)+\left(a-b\right)\left(b-c\right)\left(a-c\right)\)
\(=pq-3r-\sqrt{-4p^3r + p^2q^2 + 18pqr - 4q^3 - 27r^2}\)
\(=3p-3r-\sqrt{-4p^3r + 9p^2 + 54pr - 108 - 27r^2}\)
Do đó: \(a^2b+b^2c+c^2a\)\(=\frac{3p-3r-\sqrt{-4p^3r + 9p^2 + 54pr - 108 - 27r^2}}{2}\)
Do đó: \(P\)\(=\frac{3p-3r-\sqrt{-4p^3r + 9p^2 + 54pr - 108 - 27r^2}}{6}\)\(+\sqrt[3]{9p}\ge4\)
\(\Leftrightarrow\frac{3p-3r}{6}+\sqrt[3]{9p}\ge4+\)\(\frac{\sqrt{-4p^3r + 9p^2 + 54pr - 108 - 27r^2}}{6}\)
Or \(3p-3r+6\sqrt[3]{9p}-24\ge\)\(\sqrt{-4p^3r + 9p^2 + 54pr - 108 - 27r^2}\)
Vì: \(VT=3p-3r+6\sqrt[3]{9p}-24\ge3p-\frac{pq}{3}+18-24=0\)
Nên bất đẳng thức trên tương đương:
\(\left(3p-3r+6\sqrt[3]{9p}-24\right)^2\ge\) \(-4p^3r + 9p^2 + 54pr - 108 - 27r^2\)
Em chịu thua :( @Akai Haruma @Nguyễn Việt Lâm giúp em với ạ.
Do \(a,b< 1\Rightarrow a^3< a^2< a< 1\)
\(b^3< b^2< b< 1\)
Ta có :
\(\left(1-a^2\right)\left(1-b\right)>0\)
\(\Rightarrow1+a^2b>a^2b\)
\(\Rightarrow1+a^2b>a^3+b^3\) hay \(a^3+b^3< 1+a^2b\)
Tương tự
\(b^3+c^3< 1+b^2c\)
\(c^3+a^3< 1+c^2a\)
\(\Rightarrow2a^3+2b^3+2c^3< 3+a^2b+b^2c+c^2a\)