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\(a_1,\sqrt{x}< 7\\ \Rightarrow x< 49\\ a_2,\sqrt{2x}< 6\\ \Rightarrow x< 18\\ a_3,\sqrt{4x}\ge4\\ \Rightarrow4x\ge16\\ \Rightarrow x\ge4\\ a_4,\sqrt{x}< \sqrt{6}\\ \Rightarrow x< 6\)
\(b_1,\sqrt{x}>4\\ \Rightarrow x>16\\ b_2,\sqrt{2x}\le2\\ \Rightarrow2x\le4\\ \Rightarrow x\le2\\ b_3,\sqrt{3x}\le\sqrt{9}\\ \Rightarrow3x\le9\\ \Rightarrow x\le3\\ b_4,\sqrt{7x}\le\sqrt{35}\\ \Rightarrow7x\le35\\ \Rightarrow x\le5\)
Áp dụng bđt Cauchy ta có :
\(\sqrt{4a+1}\le\frac{4a+1+1}{2}=2a+1\)
\(\sqrt{4b+1}\le\frac{4b+1+1}{2}=2b+1\)
\(\sqrt{4c+1}\le\frac{4c+1+1}{2}=2c+1\)
\(\Rightarrow\sqrt{4a+1}+\sqrt{4b+1}+\sqrt{4b+1}\le2\left(a+b+c\right)+3=5\)(đpcm)
Áp dụng BĐT Bu-nhi-a-cốp-ski, ta có:
\(\left(1+1+1\right)\left[\left(\sqrt{4a+1}\right)^2+\left(\sqrt{4b+1}\right)^2+\left(\sqrt{4c+1}\right)^2\right]\)
\(\ge\left(\sqrt{4a+1}+\sqrt{4b+1}+\sqrt{4c+1}\right)^2\)
\(\Leftrightarrow\left(\sqrt{4a+1}+\sqrt{4b+1}+\sqrt{4c+1}\right)^2\le3\left(4a+1+4b+1+4c+1\right)\)
\(\Leftrightarrow VT^2\le21\)
\(\Rightarrow VT^2< 25\)
\(\Rightarrow VT< 5\)
Vậy \(\sqrt{4a+1}+\sqrt{4c+1}+\sqrt{4b+1}< 5\)
Đề: Cho a, b, c, d là 4 số dương thoả mãn abcd = 1. Chứng minh rằng: \(\left(\sqrt{1+a}+\sqrt{1+b}\right)\left(\sqrt{1+c}+\sqrt{1+d}\right)\ge8\)
~ ~ ~ ~ ~
Áp dụng BĐT AM - GM, ta có:
\(\left(\sqrt{1+a}+\sqrt{1+b}\right)\left(\sqrt{1+c}+\sqrt{1+d}\right)\)
\(\ge2\sqrt[4]{\left(1+a\right)\left(1+b\right)}\times2\sqrt[4]{\left(1+c\right)\left(1+d\right)}\)
\(=4\sqrt[4]{\left(1+a\right)\left(1+b\right)\left(1+c\right)\left(1+d\right)}\)
\(\ge4\sqrt[4]{2\sqrt{a}\times2\sqrt{b}\times2\sqrt{c}\times2\sqrt{d}}\)
\(=4\sqrt[4]{16\sqrt{abcd}}\)
= 8 (đpcm)
Dấu "=" xảy ra khi a = b = c = d = 1
Ỏ
Bạn tốt qá he
\(x,y \geq 0\)