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Áp dụng BĐT Cauchy-Schwarz ta có:
\(\dfrac{a}{\sqrt{a^2+1}}=\dfrac{a}{\sqrt{a^2+ab+bc+ca}}=\dfrac{a}{\sqrt{\left(a+b\right)\left(a+c\right)}}\)
\(\le\dfrac{1}{2}\left(\dfrac{a}{a+b}+\dfrac{a}{a+c}\right)\). Thiếp lập 2 BĐT còn lại:
\(\dfrac{b}{\sqrt{b^2+1}}\le\dfrac{1}{2}\left(\dfrac{b}{b+c}+\dfrac{b}{a+b}\right);\dfrac{c}{\sqrt{c^2+1}}\le\dfrac{1}{2}\left(\dfrac{c}{c+a}+\dfrac{c}{b+c}\right)\)
Cộng theo vế 3 BĐT trên ta có:
\(A\le\dfrac{1}{2}\left(\dfrac{a+b}{a+b}+\dfrac{b+c}{b+c}+\dfrac{c+a}{c+a}\right)=\dfrac{1}{2}\cdot3=\dfrac{3}{2}\)
Xảy ra khi \(a=b=c=\dfrac{1}{\sqrt{3}}\)
Áp dụng BĐT Bunyakovsky, ta có:
\(a+b+c\le\sqrt{3(a^2+b^2+c^2)}=\sqrt{3.3}=3\)
Áp dụng BĐT Cauchy, ta có:
\(A=\sum{\dfrac{1}{\sqrt{1+8a^3}}}=\sum{\dfrac{1}{\sqrt{(2a+1)(4a^2-2a+1)}}} \\\ge\sum{\dfrac{1}{\dfrac{4a^2+2}{2}}}=\sum{\dfrac{1}{2a^2+1}} \)
Ta cần chứng minh: \(\dfrac{1}{2a^2+1}\ge\dfrac{-4}{9}a+\dfrac{7}{9} \\<=>\dfrac{8a^3-14a^2+4a+2}{9(2a^2+1)}\ge0 \\<=>\dfrac{2(a-1)^2(4a+1)}{9(2a^2+1)}\ge0 (luôn\ đúng\ với\ mọi\ a>0) \\->\sum{\dfrac{1}{2a^2+1}}\ge\dfrac{-4}{9}(a+b+c)+\dfrac{21}{9}\ge\dfrac{-4}{9}.3+\dfrac{21}{9}=1 \\->A\ge1 \)
Đẳng thức xảy ra khi a = b = c = 1.
Vậy GTNN của A là 1 (khi a = b = c = 1).
Đề bài sai
Đề đúng: \(\dfrac{1}{\sqrt{a}+2\sqrt{b}+3}+\dfrac{1}{\sqrt{b}+2\sqrt{c}+3}+\dfrac{1}{\sqrt{c}+2\sqrt{a}+3}\le\dfrac{1}{2}\)
\(a^2+b^2-ab\ge\dfrac{1}{2}\left(a+b\right)^2-\dfrac{1}{4}\left(a+b\right)^2=\dfrac{1}{4}\left(a+b\right)^2\)
\(\Rightarrow\dfrac{1}{\sqrt{a^2-ab+b^2}}\le\dfrac{1}{\sqrt{\dfrac{1}{4}\left(a+b\right)^2}}=\dfrac{2}{a+b}\le\dfrac{1}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\)
Tương tự:
\(\dfrac{1}{\sqrt{b^2-bc+c^2}}\le\dfrac{1}{2}\left(\dfrac{1}{b}+\dfrac{1}{c}\right)\) ; \(\dfrac{1}{\sqrt{c^2-ca+a^2}}\le\dfrac{1}{2}\left(\dfrac{1}{c}+\dfrac{1}{a}\right)\)
Cộng vế:
\(P\le\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=3\)
Dấu "=" xảy ra khi \(a=b=c=1\)
Ta có \(3a+1\ge\left(\dfrac{\sqrt{10}-1}{3}a+1\right)^2\Leftrightarrow a\left(3-a\right)\ge0\) (luôn đúng)
Do đó \(\sqrt{3a+1}\ge\dfrac{\sqrt{10}-1}{3}a+1\).
Tương tự, \(\sqrt{3b+1}\ge\dfrac{\sqrt{10}-1}{3}b+1;\sqrt{3c+1}\ge\dfrac{\sqrt{10}-1}{3}c+1\).
Do đó \(\sqrt{3a+1}+\sqrt{3b+1}+\sqrt{3c+1}\ge\sqrt{10}+2\).
Dấu "=" xảy ra khi chẳng hạn a = 3; b = c = 0
Tham khảo:
https://hoc24.vn/hoi-dap/tim-kiem?id=219071991005&q=Cho%203%20s%E1%BB%91%20th%E1%BB%B1c%20kh%C3%B4ng%20%C3%A2m%20a%2Cb%2Cc%20v%C3%A0%20a%20b%20c%3D3%20T%C3%ACm%20GTLN%20v%C3%A0%20GTNN%20c%E1%BB%A7a%20bi%E1%BB%83u%20th%E1%BB%A9c%20K%3D%5C%28%5Csqrt%7B3a%201%7D%20%5Csqrt%7B3b%201%7D%20%5Csqrt%7B3c%201%7D%5C%29
Câu 1:
Ta có: Áp dụng BĐT phụ \(\left(x+y+z\right)\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\ge9\)
=> \(2\left(a+b+c\right)\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\ge9\)
=> \(\left(a+b+c\right)\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{a+c}\right)\ge4,5\) (*)
và BĐT Cau -chy ta có:
\(P+3=\dfrac{a+b+c}{b+c}+\dfrac{a+b+c}{a+c}+\dfrac{a+b+c}{a+b}\)
\(+\left(\dfrac{a}{b}+\dfrac{b}{a}\right)+\left(\dfrac{b}{c}+\dfrac{c}{b}\right)+\left(\dfrac{a}{c}+\dfrac{c}{a}\right)\)
<=> \(P+3\ge\left(a+b+c\right)\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{a+c}\right)\)
\(+2\sqrt{\dfrac{a}{b}.\dfrac{b}{a}}+2\sqrt{\dfrac{b}{c}.\dfrac{c}{a}}+2\sqrt{\dfrac{a}{c}.\dfrac{c}{a}}\)
<=> \(P+3\ge4,5+6=10,5\) ( Theo (*)) => \(P\ge7,5\)
=> Dấu = xảy ra <=> a = b = c
từ $x\le 3$ suy ra $x=3$ là điểm rơi
suy ra $y=8$ suy ra $P_{max}= 3*8=24$
\(=\left(1^2+4^2\right)\left(a^2+\dfrac{1}{b^2}\right)\ge\left(1a+4.\dfrac{1}{b}\right)^2\\ \Rightarrow\sqrt{a^2+\dfrac{1}{vb^2}}\ge\dfrac{1}{\sqrt{17}}\left(a+\dfrac{4}{b}\right)\)
Tương tự
\(\sqrt{b^2+\dfrac{1}{c^2}}\ge\dfrac{1}{\sqrt{17}}\left(b+\dfrac{4}{c}\right)\\ \sqrt{c^2+\dfrac{1}{a^2}}\ge\dfrac{1}{\sqrt{17}}\left(c+\dfrac{4}{a}\right)\\ Do.đó:\\ Q\ge\dfrac{1}{\sqrt{17}}\left(a+b+c+\dfrac{4}{a}+\dfrac{4}{b}+\dfrac{4}{c}\right)\ge\dfrac{1}{\sqrt{17}}\\ \left(a+b+c+\dfrac{36}{a+b+c}\right)\)
\(=\dfrac{1}{\sqrt{17}}\\ \left[a+b+c+\dfrac{9}{4\left(a+b+c\right)}+\dfrac{135}{4\left(a+b+c\right)}\right]\\ \ge\dfrac{3\sqrt{17}}{2}\)
Cái thứ nhất là tại sao có cái đầu tiên =)) cái thứ 2 dấu bằng xảy ra khi nào :V
\(A=\dfrac{x\sqrt{x}-3}{x-2\sqrt{x}-3}-\dfrac{2\left(\sqrt{x}-3\right)}{\sqrt{x}+1}+\dfrac{\sqrt{x}+3}{3-\sqrt{x}}\)
\(=\dfrac{x\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}-\dfrac{2\left(\sqrt{x}-3\right)}{\sqrt{x}+1}-\dfrac{\sqrt{x}+3}{\sqrt{x}-3}\)
\(=\dfrac{\left(x\sqrt{x}-3\right)-2\left(\sqrt{x}-3\right)^2-\left(\sqrt{x}+3\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\left(x\sqrt{x}-3\right)-\left(2x-12\sqrt{x}+18\right)-\left(x+4\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x\sqrt{x}-3x+8\sqrt{x}-24}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\left(\sqrt{x}-3\right)\left(x+8\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}=\dfrac{x+8}{\sqrt{x}+1}\)
~ ~ ~
\(\dfrac{x+8}{\sqrt{x}+1}=\dfrac{\left(4\sqrt{x}+4\right)+\left(x-4\sqrt{x}+4\right)}{\sqrt{x}+1}\)
\(=4+\dfrac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}+1}\ge4\)
Dấu "=" xảy ra khi x = 4
Áp dụng BĐT AM-GM ta có:
\(\dfrac{a^3}{\sqrt{b^2+3}}+\dfrac{a^3}{\sqrt{b^2+3}}+\dfrac{b^2+3}{7\sqrt{7}}\)
\(\ge3\sqrt[3]{\dfrac{a^3}{\sqrt{b^2+3}}\cdot\dfrac{a^3}{\sqrt{b^2+3}}\cdot\dfrac{b^2+3}{7\sqrt{7}}}=\dfrac{3a^2}{\sqrt{7}}\)
Tương tự cho 2 BĐT còn lại ta cũng có:
\(\dfrac{b^3}{\sqrt{c^2+3}}+\dfrac{b^3}{\sqrt{c^2+3}}+\dfrac{c^2+3}{7\sqrt{7}}\ge\dfrac{3b^2}{\sqrt{7}};\dfrac{c^3}{\sqrt{a^2+3}}+\dfrac{c^3}{\sqrt{a^2+3}}+\dfrac{a^2+3}{7\sqrt{7}}\ge\dfrac{3c^2}{\sqrt{7}}\)
Cộng theo vế 3 BĐT trên ta có:
\(2P+\dfrac{a^2+b^2+c^2+9}{7\sqrt{7}}\ge\dfrac{3\left(a^2+b^2+c^2\right)}{\sqrt{7}}\)
\(\Rightarrow P\ge\dfrac{\dfrac{\dfrac{\left(a+b+c\right)^2}{3}+9}{7\sqrt{7}}-\dfrac{3\cdot\dfrac{\left(a+b+c\right)^2}{3}}{\sqrt{7}}}{2}\ge\dfrac{\dfrac{\sqrt{7}}{21}}{2}=\dfrac{\sqrt{7}}{42}\)
Xảy ra khi \(a=b=c=\dfrac{1}{3}\)
am-gm :a3/V(b2+3)+a3/V(b2+3)+(b2+3)/x tự tìm số x dựa theo Min của bài (dự đoán a=b=c=1/3)