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Áp dụng cosi ta có \(a.a.a.b.b\le\frac{3a^5+2b^5}{5};b.b.b.a.a\le\frac{3b^5+2a^5}{5}\)
=> \(a^5+b^5\ge a^2b^2\left(a+b\right)\)
Khi đó
\(VT\le\frac{1}{ab\sqrt{a+b}}+\frac{1}{bc\sqrt{b+c}}+\frac{1}{ac\sqrt{a+c}}\)
Áp dụng BĐT buniacoxki ta có :
\((\frac{1}{ab\sqrt{a+b}}+\frac{1}{bc\sqrt{b+c}}+\frac{1}{ac\sqrt{a+c}})^2\le\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\left(\frac{1}{b^2\left(a+b\right)}+\frac{1}{c^2\left(b+c\right)}+...\right)\)
Mà 1/a^2+1/b^2+1/c^2=1(giả thiết)
=> \(VT\le VP\)(ĐPCM)
Dấu bằng xảy ra khi a=b=c=can(3)
\(ab+a+b=\frac{5}{4}\Rightarrow\frac{a^2+b^2}{2}+\sqrt{2\left(a^2+b^2\right)}\ge\frac{5}{4}\)
\(\Rightarrow a^2+b^2\ge\frac{1}{2}\)
\(A=\sqrt{a^4+1}+\sqrt{b^4+1}\ge\sqrt{\left(a^2+b^2\right)^2+4}\ge\sqrt{\frac{1}{4}+4}=\frac{\sqrt{17}}{2}\)
Dấu "=" xảy ra khi \(a=b=\frac{1}{2}\)
\(15\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)+30\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)=40\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)+2007\)
\(\Leftrightarrow15\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=40\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)+2007\)
\(\Leftrightarrow15\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2\le\frac{40}{3}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2+2007\)
\(\Leftrightarrow\frac{5}{3}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2\le2007\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\le\sqrt{\frac{6021}{5}}\)
Ta có:
\(5a^2+2ab+2b^2=4a^2+2ab+b^2+a^2+b^2\ge4a^2+2ab+b^2+2ab=\left(2a+b\right)^2\)
\(\Rightarrow\sqrt{5a^2+2ab+2b^2}\ge2a+b\Rightarrow\frac{1}{\sqrt{5a^2+2ab+2b^2}}\le\frac{1}{2a+b}\)
\(\Rightarrow P\le\frac{1}{2a+b}+\frac{1}{2b+c}+\frac{1}{2c+a}=\frac{1}{a+a+b}+\frac{1}{b+b+c}+\frac{1}{c+c+a}\)
\(\Rightarrow P\le\frac{1}{9}\left(\frac{1}{a}+\frac{1}{a}+\frac{1}{b}+\frac{1}{b}+\frac{1}{b}+\frac{1}{c}+\frac{1}{c}+\frac{1}{c}+\frac{1}{a}\right)=\frac{1}{3}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
\(\Rightarrow P\le\frac{1}{3}\sqrt{\frac{6021}{5}}\)
Dấu "=" xảy ra khi \(a=b=c=3\sqrt{\frac{5}{6021}}\)
Mẫu thức như vầy thì tìm max còn được chứ tìm min sao nổi bạn?
1)
\(2a+\frac{4}{a}+\frac{16}{a+2}=\left(a+\frac{4}{a}\right)+\left[\left(a+2\right)+\frac{16}{a+2}\right]-2\ge4+8-2=10\)
Dấu "=" xảy ra khi a=2
2)
\(\hept{\begin{cases}\sqrt{a\left(1-4a\right)}=\frac{1}{2}\sqrt{4a\left(1-4a\right)}\le\frac{1}{2}\cdot\frac{4a+1-4a}{2}=\frac{1}{4}\\\sqrt{b\left(1-4b\right)}=\frac{1}{2}\sqrt{4\left(1-4a\right)}\le\frac{1}{2}\cdot\frac{4b+1-4b}{2}=\frac{1}{4}\\\sqrt{c\left(1-4c\right)}=\frac{1}{2}\sqrt{4c\left(1-4c\right)}\le\frac{1}{2}\cdot\frac{4c+1-4c}{2}=\frac{1}{4}\end{cases}}\)
\(\Rightarrow\sqrt{a\left(1-4a\right)}+\sqrt{b\left(1-4b\right)}+\sqrt{c\left(1-4c\right)}\le\frac{3}{4}\)
Dấu "=" xảy ra khi \(a=b=c=\frac{1}{8}\)
Đặt \(x=\frac{1}{a}, y=\frac{1}{b}, z=\frac{1}{c}, \Rightarrow x+y+z=2\)
Suy ra \(\frac{1}{a\left(2a-1\right)^2}+\frac{1}{b\left(2b-1\right)^2}+\frac{1}{c\left(2c-1\right)^2}=\frac{x^3}{\left(2-x\right)^2}+\frac{y^3}{\left(2-y\right)^2}+\frac{z^3}{\left(2-z\right)^2}\)
Ta có \(\frac{x^3}{\left(2-x\right)^2}+\frac{2-x}{8}+\frac{2-x}{8}\ge3\sqrt[3]{\frac{x^3}{\left(2-x\right)^2} .\frac{2-x}{8}.\frac{2-x}{8}}=\frac{3x}{4}.\)
\(\Rightarrow\frac{x^3}{\left(2-x\right)^2}\ge x-\frac{1}{2}\)\(\Rightarrow\frac{x^3}{\left(2-x\right)^2}+\frac{y^3}{\left(2-y\right)^2}+\frac{z^3}{\left(2-z\right)^2}\ge x+y+z-\frac{3}{2}=2-\frac{3}{2}=\frac{1}{2}\)
dấu "=" xảy ra khi \(x=y=z=\frac{2}{3}\)hay \(a=b=c=\frac{3}{2}\)
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.
\(P^2=\left(9+a^2b^2\right)\left(\frac{1}{a}+\frac{1}{b}\right)^2=\left(\frac{3}{a}+\frac{3}{b}\right)^2+\left(a+b\right)^2\)
\(P^2\ge\left(\frac{12}{a+b}\right)^2+\left(a+b\right)^2=\frac{144}{\left(a+b\right)^2}+\frac{9\left(a+b\right)^2}{16}+\frac{7\left(a+b\right)^2}{16}\)
\(P^2\ge2\sqrt{\frac{144.9}{16}}+\frac{7.4^2}{16}=25\)
\(\Rightarrow P\ge5\)
Đặt P=\(\sqrt{9+a^2b^2}\left(\frac{1}{a}+\frac{1}{b}\right)\)
\(=\sqrt{9\left(\frac{1}{a}+\frac{1}{b}\right)^2+a^2b^2\left(\frac{1}{a}+\frac{1}{b}\right)^2}\)
\(=\sqrt{\left(\frac{3}{a}+\frac{3}{b}\right)^2+\left(a+b\right)^2}\)
Theo cauchy-schwartz:
\(\left(\left(\frac{3}{a}+\frac{3}{b}\right)^2+\left(a+b\right)^2\right)\left(\left(\frac{3}{4}\right)^2+1^2\right)\ge\left[\frac{9}{4}\left(\frac{1}{a}+\frac{1}{b}\right)+a+b\right]^2\)
\(\frac{9}{4}\left(\frac{1}{a}+\frac{1}{b}\right)+a+b\ge\frac{9}{4}.\frac{4}{a+b}+a+b=\frac{9}{a+b}+a+b\)
Theo AM-GM:
\(\frac{9}{a+b}+a+b=a+b+\frac{16}{a+b}-\frac{7}{a+b}\ge2\sqrt{\left(a+b\right)\frac{16}{a+b}}-\frac{7}{a+b}\)
Mà a+b≥4
\(\Rightarrow\frac{9}{a+b}+a+b\ge2\sqrt{16}-\frac{7}{4}=\frac{25}{4}\)
=>P2≥\(\frac{\left(\frac{25}{4}\right)^2}{\left(\frac{3}{4}\right)^2+1^2}=5^2\)
=>P≥5
Dấu bằng xảy ra khi a=b=2
Vậy minP=5 khi a=b=2