Inequalities Involving A.M, G.M, H.M

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

\(A.M=\frac{{{4}^{{{\sin }^{2}}x}}+{{4}^{{{\cos }^{2}}x}}}{2}\),

\(G.M=\sqrt{{{4}^{{{\sin }^{2}}x}}\times {{4}^{{{\cos }^{2}}x}}}\),

\(\frac{{{4}^{{{\sin }^{2}}x}}+{{4}^{{{\cos }^{2}}x}}}{2}=\sqrt{{{4}^{{{\sin }^{2}}x}}\times {{4}^{{{\cos }^{2}}x}}}\),

\({{4}^{{{\sin }^{2}}x}}+{{4}^{{{\cos }^{2}}x}}=2\sqrt{{{4}^{{{\sin }^{2}}x}}\times {{4}^{{{\cos }^{2}}x}}}\),

\({{4}^{{{\sin }^{2}}x}}+{{4}^{{{\cos }^{2}}x}}=2\sqrt{{{4}^{{{\sin }^{2}}x+{{\cos }^{2}}x}}}\),

\({{4}^{{{\sin }^{2}}x}}+{{4}^{{{\cos }^{2}}x}}=2\sqrt{{{4}^{\left( 1 \right)}}}\),

\({{4}^{{{\sin }^{2}}x}}+{{4}^{{{\cos }^{2}}x}}=2\times 2\),

\({{4}^{{{\sin }^{2}}x}}+{{4}^{{{\cos }^{2}}x}}=4\),

\({{\left( {{4}^{{{\sin }^{2}}x}}+{{4}^{{{\cos }^{2}}x}} \right)}_{\min }}=4\).