Inequalities Involving A.M, G.M, H.M
Let A, G and H be Arithmetic, Geometric, and Harmonic means of two positive numbers a and b.
Then, \(A=\frac{a+b}{2}\), \(G=\sqrt{ab}\) and \(H=\frac{2ab}{a+b}\).
These three means possess the following property.
A ≥ G ≥ H.
Proof: We have \(A=\frac{a+b}{2}\), \(G=\sqrt{ab}\) and \(H=\frac{2ab}{a+b}\).
\(A-G=\frac{a+b}{2}-\sqrt{ab}\),
\(=\frac{a+b-2\sqrt{ab}}{2}\),
\(=\frac{{{\left( \sqrt{a}-\sqrt{b} \right)}^{2}}}{2}\),
\(=\frac{{{\left( \sqrt{a}-\sqrt{b} \right)}^{2}}}{2}\ge 0\),
A ≥ G … (1)
\(G-H=\sqrt{ab}-\frac{2ab}{a+b}\),
\(G-H=\frac{-2ab+\left( a+b \right)\sqrt{ab}}{a+b}\),
\(G-H=\sqrt{ab}\left( \frac{\left( a+b \right)-2\sqrt{ab}}{a+b} \right)\),
\(=\frac{\sqrt{ab}}{a+b}{{\left( \sqrt{a}-\sqrt{b} \right)}^{2}}\ge 0\),
G ≥ H … (2)
From Equation (1) and (2), we will get
A ≥ G ≥ H.
Example: Find the minimum value of \({{4}^{{{\sin }^{2}}x}}+{{4}^{{{\cos }^{2}}x}}\).
Solution: AM ≥ GM
\(A.M=\frac{{{4}^{{{\sin }^{2}}x}}+{{4}^{{{\cos }^{2}}x}}}{2}\),
\(G.M=\sqrt{{{4}^{{{\sin }^{2}}x}}\times {{4}^{{{\cos }^{2}}x}}}\),
\(\frac{{{4}^{{{\sin }^{2}}x}}+{{4}^{{{\cos }^{2}}x}}}{2}=\sqrt{{{4}^{{{\sin }^{2}}x}}\times {{4}^{{{\cos }^{2}}x}}}\),
\({{4}^{{{\sin }^{2}}x}}+{{4}^{{{\cos }^{2}}x}}=2\sqrt{{{4}^{{{\sin }^{2}}x}}\times {{4}^{{{\cos }^{2}}x}}}\),
\({{4}^{{{\sin }^{2}}x}}+{{4}^{{{\cos }^{2}}x}}=2\sqrt{{{4}^{{{\sin }^{2}}x+{{\cos }^{2}}x}}}\),
\({{4}^{{{\sin }^{2}}x}}+{{4}^{{{\cos }^{2}}x}}=2\sqrt{{{4}^{\left( 1 \right)}}}\),
\({{4}^{{{\sin }^{2}}x}}+{{4}^{{{\cos }^{2}}x}}=2\times 2\),
\({{4}^{{{\sin }^{2}}x}}+{{4}^{{{\cos }^{2}}x}}=4\),
\({{\left( {{4}^{{{\sin }^{2}}x}}+{{4}^{{{\cos }^{2}}x}} \right)}_{\min }}=4\).