Cho \(a,b\ge1\)
\(CMR:a\sqrt{b-1}+b\sqrt{a-1}\le ab\)
Mong các cao nhân ra tay giúp đỡ ạ,nhớ áp dụng bđt Côsi nha
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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
\(a\sqrt{b-1}+b\sqrt{a-1}\Leftrightarrow\sqrt{a}\sqrt{ab-a}+\sqrt{b}\sqrt{ab-b}\)
\(\le\sqrt{\left(a+b\right)\left(2ab-a-b\right)}\le\frac{a+b-a-b+2ab}{2}=ab\)
BĐT đc chứng minh
\(x=\sqrt{a-1};y=\sqrt{b-1}\) bỏ căn đi viết cho dẽ nhìn
\(x^2=a-1;y^2=b-1\Leftrightarrow\left(x^2+1\right)y+\left(y^2+1\right)x\le\left(x^2+1\right)\left(y^2+1\right)\)
\(\Leftrightarrow\left(x^2+1\right)\left(y^2-2y+1\right)+\left(y^2+1\right)\left(x^2-2x+1\right)\ge0\)
\(\Leftrightarrow\left(x^2+1\right)\left(y-1\right)^2+\left(y^2+1\right)\left(x-1\right)^2\ge0\)Đúng với mọi x,y => dpcm
Đẳng thức khi x=y=1=> a=b=2
\(\left(2+7\right)\left(2a^2+\dfrac{7}{b^2}\right)\ge\left(2a+\dfrac{7}{b}\right)^2\)
\(\Rightarrow\sqrt{2a^2+\dfrac{7}{b^2}}\ge\dfrac{1}{3}\left(2a+\dfrac{7}{b}\right)\)
Tương tự: \(\sqrt{2b^2+\dfrac{7}{c^2}}\ge\dfrac{1}{3}\left(2a+\dfrac{7}{c}\right)\) ; \(\sqrt{2c^2+\dfrac{7}{a^2}}\ge\dfrac{1}{3}\left(2c+\dfrac{7}{a}\right)\)
Cộng vế:
\(VT\ge\dfrac{1}{3}\left(2a+2b+2c+\dfrac{7}{a}+\dfrac{7}{b}+\dfrac{7}{c}\right)=2+\dfrac{7}{3}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
\(VT\ge2+\dfrac{7}{9}.\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\) (do \(a+b+c=3\))
\(VT\ge2+\dfrac{7}{9}.\left(\sqrt{a}.\sqrt{\dfrac{1}{a}}+\sqrt{b}.\sqrt{\dfrac{1}{b}}+\sqrt{c}.\sqrt{\dfrac{1}{c}}\right)^2=9\) (đpcm)
Dấu "=" xảy ra khi \(a=b=c=1\)
a p dg côsi \(a\sqrt{b-1}=a.1.\sqrt{b-1}\le a.\dfrac{1+b-1}{2}=\dfrac{ab}{2}\)
ttuong tu \(b\sqrt{a-1}\le\dfrac{ab}{2}\)
nên vt\(\le ab\)
dau = xảy ra a=b=2
\(\dfrac{a}{a+2\sqrt{\left(a+bc\right)}}=\dfrac{a}{a+2\sqrt{a\left(a+b+c\right)+bc}}=\dfrac{a}{a+2\sqrt{\left(a+b\right)\left(a+c\right)}}\)
\(=\dfrac{a}{a+\dfrac{\sqrt{\left(a+b\right)\left(a+c\right)}}{2}+\dfrac{\sqrt{\left(a+b\right)\left(a+c\right)}}{2}+\dfrac{\sqrt{\left(a+b\right)\left(a+c\right)}}{2}+\dfrac{\sqrt{\left(a+b\right)\left(a+c\right)}}{2}}\)
\(\le\dfrac{a}{5^2}\left(\dfrac{1}{a}+\dfrac{1}{\dfrac{\sqrt{\left(a+b\right)\left(a+c\right)}}{2}}+\dfrac{1}{\dfrac{\sqrt{\left(a+b\right)\left(a+c\right)}}{2}}+\dfrac{1}{\dfrac{\sqrt{\left(a+b\right)\left(a+c\right)}}{2}}+\dfrac{1}{\dfrac{\sqrt{\left(a+b\right)\left(a+c\right)}}{2}}\right)\)
\(=\dfrac{a}{25}\left(\dfrac{1}{a}+\dfrac{8}{\sqrt{\left(a+b\right)\left(a+c\right)}}\right)=\dfrac{1}{25}+\dfrac{8}{25}.\dfrac{a}{\sqrt{\left(a+b\right)\left(a+c\right)}}\)
\(\le\dfrac{1}{25}+\dfrac{4}{25}\left(\dfrac{a}{a+b}+\dfrac{a}{a+c}\right)\)
Tương tự:
\(\dfrac{b}{b+2\sqrt{b+ac}}\le\dfrac{1}{25}+\dfrac{4}{25}\left(\dfrac{b}{a+b}+\dfrac{b}{b+c}\right)\)
\(\dfrac{c}{c+2\sqrt{c+ab}}\le\dfrac{1}{25}+\dfrac{4}{25}\left(\dfrac{c}{a+c}+\dfrac{c}{b+c}\right)\)
Cộng vế:
\(P\le\dfrac{3}{25}+\dfrac{4}{25}\left(\dfrac{a+b}{a+b}+\dfrac{b+c}{b+c}+\dfrac{c+a}{c+a}\right)=\dfrac{15}{25}=\dfrac{3}{5}\)
Dấu "=" xảy ra khi \(a=b=c=\dfrac{1}{3}\)
a√(b-1) = a√1(b-1) ≤ b/2*a=ab/2
b√(a-1) = b√1(a-1) ≤ a/2*b=ab/2
Cộng vế theo vế ta được:
a√(b-1) + b√(a-1) ≤ ab/2 +ab/2 = 2ab/2 = ab
a√(b-1) = a√1(b-1) ≤ b/2*a=ab/2
b√(a-1) = b√1(a-1) ≤ a/2*b=ab/2
Cộng vế theo vế ta được:
a√(b-1) + b√(a-1) ≤ ab/2 +ab/2 = 2ab/2 = ab